We know that for a relation to be an "equivalence relation" it should satisfy three different types of relations: Transitive, Reflexive, Symmetric
Let's notch this down from Graphs to sets, to make things a bit easier. I will give two examples, one for equivalence relation and another for not equivalence relation
Assume, A = { 1,2,3 } and B = { 1, 2, 3 }
Cartesian Product: A X B = { ( 1,1 ) (1,2) (1,3) (2,1) (2,2) (2,3) ( 3,1 ) (3,2) (3,3) }
Let's say that the Relation R = Cartesian Product
We can see that this Relation R is indeed an "equivalence relation"
For a counter-example, we can think of the sets as A = { 1,2,3 } and B = { 4,5 }.
Cartesian Product: A X B and Relation R = Cartesian Product
In this case, we know that any Relation R would, at least, fail to satisfy the "Reflexive property" and hence it would not be an "equivalence relation"
Now, that we are clear ~ We can bounce back to our original problem of "Proving Isomorphism Relation in Graphs is an Equivalence Relation"
Let's walk through this, bit by bit:
In general, we know that two graphs are isomorphic when they are structurally the same ( irrespective of how we draw them )
For a moment, if we think of "isomorphism" as a function ~ we know it's range is "True or False", meaning this function is what we call a "property" or "predicate"
We also know that, when a predicate has a domain consisting "set of k-tuples", it becomes a relation
My question is, when we say "isomorphism is an equivalence relation" at what level are we speaking?
Are we talking with respect to the vertex set of the Graphs? or are we talking at a more zoomed out level thinking of Graphs as a whole?
Specifically,
Assume G1 and G2 are isomorphic, G1 = (V1, E1 ) and G2 =(V2, E2)
V1 = (a, b, c, d) E1 = {(a,d) (b,c) (b,d) (c,b) (c,d) (d,a)(d,b)(d,c)}
V2 = (e, f, g, h) E2 = {(e,h) (f,g) (f,h) (g,f) (g,h) (h,e)(h,f)(h,g)}
So, when we say G1 and G2 are symmetric, do we say that ~ okay the cartesian product of V1 and V2 is {..} and here is this relation from this cartesian product {..} ~ which won't be an equivalence relation because the two sets are different so since it's an isomophic relation, do we assume that G2 has the same ordered pair ( V2, E2 ) as G1?
or do we look at it from a more zoomed out level, and go like "okay, we know that these two are isomorphic to each other meaning that some function is mapping the vertices of G1 to G2 while following some rules and since it's a bijection function we can inverse it and still get a proper bijection, hence it must be symmetric" <- so, do we just use "general concepts" of reflexive, transitive and symmetry without thinking at the set level but more at a logical /object level like if a-> b and b->a then it's symmetrix
I’m not sure that I understand your question correctly but if you want to prove that graph isomorphism is an equivalence relation you can do that by showing that it has the required properties (which you listed yourself). For example you can break down isomorphism into two bijective mapping for vertices and edges and study their properties.
I see you mentioning cartesian product few times, and I guess that you're also wondering about some sort of an explicit construction of a binary relation for two graphs with different underlying vertices sets? In that case, given graphs $G$ and $H$ with $V(G) \subset \mathbb{N}$, $V(H) \subset \mathbb{Q}$ you can construct $G^*$ and $H^*$ — sets of all possible graphs with vertices in $\mathbb{N}$ and $\mathbb{Q}$ respectively. Then you can define a binary relation as: $G \sim H \Leftrightarrow (G, H) \in G^* \times H^*$.