# Surjectivity Foundations

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261004-advance-20261005-r02-surjective-foundations · 2026-10-04 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. Surjectivity of a Finite Total Function

The declared total function f:A→B has A=∅, B=∅, and table ∅. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0001/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 2. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0002/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 3. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0003/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 4. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0004/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 5. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0005/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 6. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0006/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 7. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0007/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 8. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0008/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 9. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0009/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 10. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0010/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 11. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0011/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 12. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0012/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 13. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0013/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 14. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0014/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 15. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0015/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 16. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0016/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 17. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0017/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 18. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0018/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 19. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0019/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 20. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0020/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 21. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0021/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 22. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0022/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 23. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0023/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 24. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f surjective onto the declared codomain B?

View answer and explanationYes

Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0024/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 25. Surjectivity of a Finite Total Function

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. Is f surjective onto the declared codomain B?

View answer and explanationNo

Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0025/)

Surjectivity and codomain coverage · qf-surjective-table-property · candidate





## 26. Uncovered Codomain Targets

The declared total function f:A→B has A=∅, B=∅, and table ∅. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0026/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 27. Uncovered Codomain Targets

The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0027/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 28. Uncovered Codomain Targets

The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{5}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {5}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0028/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 29. Uncovered Codomain Targets

The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0029/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 30. Uncovered Codomain Targets

The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{-3, -2}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-3, -2}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0030/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 31. Uncovered Codomain Targets

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0031/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 32. Uncovered Codomain Targets

The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{7}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {7}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0032/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 33. Uncovered Codomain Targets

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0033/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 34. Uncovered Codomain Targets

The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{-4, -3}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-4, -3}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0034/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 35. Uncovered Codomain Targets

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0035/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 36. Uncovered Codomain Targets

The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{9}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {9}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0036/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 37. Uncovered Codomain Targets

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0037/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 38. Uncovered Codomain Targets

The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{-5, -4, -3}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-5, -4, -3}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0038/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 39. Uncovered Codomain Targets

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0039/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 40. Uncovered Codomain Targets

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{11}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {11}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0040/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 41. Uncovered Codomain Targets

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0041/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 42. Uncovered Codomain Targets

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{-6, -5, -4}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-6, -5, -4}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0042/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 43. Uncovered Codomain Targets

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0043/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 44. Uncovered Codomain Targets

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{13}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {13}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0044/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 45. Uncovered Codomain Targets

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0045/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 46. Uncovered Codomain Targets

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{-7, -6, -5, -4}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-7, -6, -5, -4}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0046/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 47. Uncovered Codomain Targets

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0047/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 48. Uncovered Codomain Targets

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{15}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {15}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0048/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 49. Uncovered Codomain Targets

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation∅

Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0049/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 50. Uncovered Codomain Targets

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.

View answer and explanation{-8, -7, -6, -5}

Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-8, -7, -6, -5}; any missing target defeats surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0050/)

Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate





## 51. Least Permitted Preimage of a Target

The declared total function f:A→B has A=∅, B={0}, and table ∅. For t=0∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0051/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 52. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. For t=1∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0052/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 53. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. For t=5∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0053/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 54. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. For t=0∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0054/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 55. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. For t=-2∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0055/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 56. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. For t=2∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0056/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 57. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. For t=7∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0057/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 58. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. For t=1∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0058/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 59. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. For t=-4∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0059/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 60. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. For t=1∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-3

Enumerate the full fiber and select its least input if it is nonempty. The answer is -3. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0060/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 61. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. For t=9∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0061/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 62. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. For t=2∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0062/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 63. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. For t=-3∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0063/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 64. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. For t=3∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-2

Enumerate the full fiber and select its least input if it is nonempty. The answer is -2. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0064/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 65. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. For t=11∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0065/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 66. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. For t=3∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0066/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 67. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. For t=-2∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-2

Enumerate the full fiber and select its least input if it is nonempty. The answer is -2. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0067/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 68. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. For t=5∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0068/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 69. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. For t=13∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0069/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 70. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. For t=4∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0070/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 71. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. For t=-1∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-5

Enumerate the full fiber and select its least input if it is nonempty. The answer is -5. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0071/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 72. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. For t=0∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation0

Enumerate the full fiber and select its least input if it is nonempty. The answer is 0. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0072/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 73. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. For t=15∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0073/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 74. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. For t=5∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanation-1

Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0074/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 75. Least Permitted Preimage of a Target

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. For t=-8∈B, find its numerically least permitted preimage, answering none when there is no preimage.

View answer and explanationnone

Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0075/)

Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate





## 76. Coverage after a Valid Codomain Change

The declared total function f:A→B has A=∅, B=∅, and table ∅. The new codomain B2=∅ contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0076/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 77. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. The new codomain B2={1, 0, 101} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0077/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 78. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. The new codomain B2={1, 3} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0078/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 79. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. The new codomain B2={0, 103} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0079/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 80. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. The new codomain B2={-1} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0080/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 81. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. The new codomain B2={1, 2, 0, 105} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0081/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 82. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. The new codomain B2={1, 3, 5} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0082/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 83. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. The new codomain B2={0, 1, 107} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0083/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 84. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. The new codomain B2={-1, -2} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0084/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 85. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. The new codomain B2={1, 2, 3, 0, 109} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0085/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 86. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. The new codomain B2={1, 3, 5, 7} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0086/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 87. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. The new codomain B2={0, 1, 2, 111} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0087/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 88. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. The new codomain B2={-1, -2} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0088/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 89. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. The new codomain B2={1, 2, 3, 4, 0, 113} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0089/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 90. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. The new codomain B2={1, 3, 5, 7, 9} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0090/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 91. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. The new codomain B2={0, 1, 2, 3, 115} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0091/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 92. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. The new codomain B2={-1, -2, -3} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0092/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 93. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. The new codomain B2={1, 2, 3, 4, 5, 0, 117} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0093/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 94. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. The new codomain B2={1, 3, 5, 7, 9, 11} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0094/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 95. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. The new codomain B2={0, 1, 2, 3, 4, 119} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0095/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 96. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. The new codomain B2={-1, -2, -3} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0096/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 97. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. The new codomain B2={1, 2, 3, 4, 5, 6, 0, 121} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0097/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 98. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. The new codomain B2={1, 3, 5, 7, 9, 11, 13} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0098/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 99. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. The new codomain B2={0, 1, 2, 3, 4, 5, 123} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationNo

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0099/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate





## 100. Coverage after a Valid Codomain Change

The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. The new codomain B2={-1, -2, -3, -4} contains every attained output. Is the same output table surjective as A→B2?

View answer and explanationYes

Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r02-surjective-0100/)

Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261004-advance-20261005-r02-surjective-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261004-advance-20261005-r02-surjective-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261004-advance-20261005-r02-surjective-foundations/validation-report.json)