Goal and scope
G has vertices 1, 2, 3, 4 and edges {1,2}, {2,3}, {3,4}, {1,4}; H has vertices a, b, c, d and edges {a,c}, {b,c}, {b,d}, {a,d}. Prove that f: 1↦a, 2↦c, 3↦b, 4↦d is a graph isomorphism G→H; show that the bijection g: 1↦a, 2↦b, 3↦c, 4↦d is not an isomorphism; and prove that the path P (edges {1,2}, {2,3}, {3,4}) is not isomorphic to H, even though f sends every edge of P to an edge of H.
Stays inside the MKO: complete edge sets of finite undirected simple graphs and a complete mapping table (assumption a1). The guide checks both requirements of an isomorphism item by item and proves in general the lemma that an isomorphism induces a bijection between edge sets, using it as an invariant for refutation; a matching degree multiset is treated only as a necessary clue (assumption a2). The general theory of walks, paths, connected components and colouring is not treated, and the canonical MKO and its English edition are not changed.
Main derivation
S01 · definition
Before
f:V(G)\to V(H)\ \text{is an isomorphism}After
f\ \text{bijective}\ \land\ \forall u\neq v:\ (\{u,v\}\in E(G)\Leftrightarrow\{f(u),f(v)\}\in E(H))Conditions: none
Rule: mko_iso_def · Uses: G1, G2, G3
Leaf basis: mko_iso_def + def_unfold
Unfold the definition of the MKO: both requirements must hold. First, f is a bijection; second, for every pair of distinct vertices, 'adjacent in G' and 'images adjacent in H' are true or false together. The second requirement is an iff: edges must go to edges and nonedges to nonedges.
S02 · implies
Before
f(1)=a,\ f(2)=c,\ f(3)=b,\ f(4)=d
After
f:V(G)\to V(H)\ \text{is a function}Conditions: none
Rule: function_def · Uses: G1, G2, G3
Leaf basis: function_def
The table gives exactly one value to each of the four vertices of V(G), and every value lies in V(H), so f is a function from V(G) to V(H). The table is complete (assumption a1), with no undefined vertex.
S03 · implies
Before
f(1),f(2),f(3),f(4)=a,c,b,d
After
f(u)=f(v)\Rightarrow u=v
Conditions: none
Rule: injective_surjective_def · Uses: S02
Leaf basis: injective_surjective_def + finite_forall
The four images a, c, b, d are pairwise distinct (C(4,2)=6 comparisons), so distinct vertices have distinct images; equivalently f(u)=f(v) implies u=v, and f is injective.
S04 · implies
Before
V(H)=\{a,b,c,d\}After
a=f(1),\ b=f(3),\ c=f(2),\ d=f(4)
Conditions: none
Rule: injective_surjective_def · Uses: S02, G2
Leaf basis: injective_surjective_def + exists_intro + finite_forall
Find a preimage for every element of V(H): a=f(1), b=f(3), c=f(2), d=f(4). So f is surjective.
S05 · implies
Before
f\ \text{injective},\ f\ \text{surjective}After
f\ \text{bijective};\ f^{-1}:\ a\mapsto 1,\ b\mapsto 3,\ c\mapsto 2,\ d\mapsto 4Conditions: none
Rule: and_intro_elim · Uses: S03, S04
Leaf basis: and_intro_elim + injective_surjective_def
Injective and surjective means bijective, with a unique inverse f⁻¹ read off the table backwards. The first requirement holds.
S06 · equivalent
Before
\forall u\neq v\in V(G):\ (\{u,v\}\in E(G)\Leftrightarrow\{f(u),f(v)\}\in E(H))After
\text{the iff for each of the}\ \tbinom42=6\ \text{unordered pairs}Conditions: none
Rule: finite_forall · Uses: S01, G1, finite_forall
Leaf basis: finite_forall + forall_elim + finite_counting_basics
The condition for (u,v) and for (v,u) is the same statement, since {u,v}={v,u} and {f(u),f(v)}={f(v),f(u)}. The ordered pairs of distinct vertices are exactly the 6 unordered pairs taken in both orders, so the universal condition is equivalent to its holding for each of the 6 unordered pairs {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}.
S07 · equivalent
Before
\{u,v\}\mapsto\{f(u),f(v)\}After
\{1,2\}\mapsto\{a,c\},\ \{1,3\}\mapsto\{a,b\},\ \{1,4\}\mapsto\{a,d\},\ \{2,3\}\mapsto\{b,c\},\ \{2,4\}\mapsto\{c,d\},\ \{3,4\}\mapsto\{b,d\}Conditions: none
Rule: function_def · Uses: S06, G3
Leaf basis: function_def + eq_replace
Look up the image of each of the 6 unordered pairs. For example the image of {2,3} is {f(2),f(3)}={c,b}={b,c}. Since f is injective, each image is again a set of two distinct vertices.
S08 · case_split
Before
\{u,v\}\in E(G)\ ?After
\{1,2\},\{2,3\},\{3,4\},\{1,4\}\in E(G);\ \{1,3\},\{2,4\}\notin E(G)Conditions: none
Rule: case_analysis · Uses: G1, S06
Leaf basis: case_analysis + graph_basics
Split the 6 unordered pairs by whether they are edges of G: four edges and two nonedges. Since E(G) is listed completely, an unlisted pair is a nonedge.
S09 · implies
Before
\{f(u),f(v)\}\in E(H)\ ?After
\{a,c\},\{b,c\},\{b,d\},\{a,d\}\in E(H);\ \{a,b\},\{c,d\}\notin E(H)Conditions: none
Rule: graph_basics · Uses: G2, S07
Leaf basis: graph_basics + set_operations_def
Look up the 6 images of S07 in the complete edge list of H: {a,c}, {b,c}, {b,d}, {a,d} are listed; {a,b} and {c,d} are not, so they are nonedges of H.
S10 · implies
Before
\text{edge status of}\ \{u,v\}\ \text{in}\ G\ \text{and of its image in}\ HAfter
\{u,v\}\in E(G)\Leftrightarrow\{f(u),f(v)\}\in E(H)\ \text{for all 6 pairs}Conditions: none
Rule: iff_intro · Uses: S07, S08, S09
Leaf basis: iff_intro + case_analysis + finite_forall
Compare pair by pair: the images {a,c}, {b,c}, {b,d}, {a,d} of the four edges {1,2}, {2,3}, {3,4}, {1,4} of G are edges of H; the images {a,b}, {c,d} of the two nonedges {1,3}, {2,4} of G are nonedges of H. For every pair both sides are true together or false together, so each iff holds.
S11 · implies
Before
f\ \text{bijective},\ \text{iff on all 6 pairs}After
f:G\to H\ \text{is an isomorphism},\ G\cong HConditions: none
Rule: mko_iso_def · Uses: S01, S05, S06, S10
Leaf basis: mko_iso_def + and_intro_elim
Bijectivity comes from S05; S06 reduces the universal adjacency condition to 6 pairs, and S10 establishes all 6. By the definition in S01, f is an isomorphism, so G≅H. The conclusion depends only on the complete mapping table and complete edge lists, not on degrees or how the figures look.
S12 · implies
Before
g(1)=a,\ g(2)=b,\ \{1,2\}\in E(G)After
\{g(1),g(2)\}=\{a,b\}\notin E(H)\Rightarrow g\ \text{not an isomorphism}Conditions: none
Rule: contradiction · Uses: S01, G1, G2, G4
Leaf basis: contradiction + quantifier_negation + injective_surjective_def
The four images a, b, c, d of g are distinct and cover V(H), so g is also a bijection. But {1,2} is an edge of G while its image {a,b} is not an edge of H, so the iff fails at this pair. One counterexample refutes the universal condition, so g is not an isomorphism. Bijectivity alone is not enough.
S13 · implies
Before
f\ \text{on}\ P:\ \{1,2\},\{2,3\},\{3,4\}\mapsto\{a,c\},\{b,c\},\{b,d\}After
\text{every edge of}\ P\ \text{goes to an edge of}\ HConditions: none
Rule: graph_basics · Uses: G4, G3, G2
Leaf basis: graph_basics + function_def
Regard f as a bijection V(P)→V(H): the images {a,c}, {b,c}, {b,d} of the three edges of P all lie in E(H). So f 'preserves edges'. This is exactly the situation the MKO warns about: preserving edges is not enough.
S14 · implies
Before
\{1,4\}\notin E(P)After
\{f(1),f(4)\}=\{a,d\}\in E(H)\Rightarrow f\ \text{is not an isomorphism}\ P\to HConditions: none
Rule: contradiction · Uses: S13, G4, G2
Leaf basis: contradiction + quantifier_negation
{1,4} is not an edge of P (the edge list of P is complete), but its image {a,d} is an edge of H. The '⇐' direction of the iff fails, so f is not an isomorphism P→H. Declaring an isomorphism after checking only that edges go to edges is wrong.
S15 · implies
Before
|E(P)|=3,\ |E(H)|=4
After
\text{no isomorphism}\ P\to H\ \text{exists}Conditions: 3\neq 4
Rule: L_edge_bij · Uses: G2, G4
Leaf basis: L_edge_bij + contrapositive + numeric_evaluation
By Lemma L_edge_bij (proved in S15.1 to S15.3), any isomorphism φ:P→H gives a bijection E(P)→E(H), so |E(P)|=|E(H)|. But 3≠4, so by contraposition there is no isomorphism P→H at all, not merely that f fails. A mismatch of the edge-count invariant suffices to refute.
S16 · implies
Before
\deg\ \text{in}\ G\ \text{and}\ HAfter
\{2,2,2,2\}=\{2,2,2,2\}Conditions: none
Rule: graph_basics · Uses: G1, G2, S11
Leaf basis: graph_basics + numeric_evaluation
In G, N(1)={2,4}, N(2)={1,3}, N(3)={2,4}, N(4)={1,3}; in H, N(a)={c,d}, N(b)={c,d}, N(c)={a,b}, N(d)={a,b}; every degree is 2, and the handshake lemma checks 4·2=2·4. The matching degree multiset is consistent with S11. In general it is only necessary: C6 and two triangles refute general sufficiency. Here, however, a simple graph on four vertices with all degrees 2 is necessarily C4, so this special condition forces both graphs to be C4. This guide instead selects the complete map proof of S11 and does not use that classification route.