Concept guide
Bijectivity: testing given functions by fiber profiles, kept apart from whether some bijection exists
Let A={1,2,3}, B={4,5,6}, and let f, g:A→B be given by tables: f(1)=5, f(2)=6, f(3)=4; g(1)=4, g(2)=4, g(3)=6. First prove, for any total function, the equivalence: bijective ⇔ the fiber of every b∈B has exactly one element. Then list the complete fiber profiles of f and g in the codomain order 4, 5, 6, and decide that f is bijective while g is neither injective nor surjective. Finally answer the different question of whether some bijection exists: one exists A→B (f is one), none exists {1,2,3}→{4,5}, and ∅→∅ has exactly one, the empty bijection.