# Graph Isomorphism is Equivalence Relation

## Contents

## Theorem

Graph isomorphism is an equivalence relation.

## Proof 1

From the formal definitions:

### Simple Graph

A (simple) graph $G$ is a non-empty set $V$ together with an antireflexive, symmetric relation $\RR$ on $V$.

### Digraph

A digraph $D$ is a non-empty set $V$ together with an antireflexive relation $\RR$ on $V$.

### Loop-graph

A loop-graph $P$ is a non-empty set $V$ together with a symmetric relation $\RR$ on $V$.

### Loop-Digraph

A loop-digraph $D$ is a non-empty set $V$ together with a relation $\RR$ on $V$.

It can be seen from these definitions that all the above are relational structures with more or less restriction on the relation.

Hence the result from Relation Isomorphism is Equivalence Relation.

$\blacksquare$

## Proof 2

In the following, let:

\(\displaystyle G_1\) | \(=\) | \(\displaystyle \struct {\map V {G_1}, \map E {G_1} }\) | |||||||||||

\(\displaystyle G_2\) | \(=\) | \(\displaystyle \struct {\map V {G_2}, \map E {G_2} }\) | |||||||||||

\(\displaystyle G_3\) | \(=\) | \(\displaystyle \struct {\map V {G_3}, \map E {G_3} }\) |

be arbitrary graphs.

Checking in turn each of the criteria for equivalence:

### Reflexivity

The identity mapping on $\map V {G_1}$ is a bijection that maps each vertex to itself.

This is seen to preserve adjacency.

Thus graph isomorphism preserves reflexivity.

$\Box$

### Symmetry

Let $G_1$ be isomorphic to $G_2$

Then there exists an isomorphism $\phi: \map V {G_1} \to \map V {G_2}$ from $\map V {G_1}$ to $\map V {G_2}$.

Let its inverse $\phi^{-1}: \map V {G_2} \to \map V {G_1}$ be defined by:

- $\map {\phi^{-1} } {v_2} = v_1 \iff \map \phi {v_1} = v_2$

By Inverse of Bijection is Bijection, $\phi^{-1}$ is a bijection.

Let $u_2, v_2 \in \map V {G_2}$ such that $\map {\phi^{-1} } {u_2} = u_1$ and $\map {\phi^{-1} } {v_2} = v_1$.

Then:

\(\displaystyle \map \phi {u_1}\) | \(=\) | \(\displaystyle u_2\) | |||||||||||

\(\displaystyle \map \phi {v_1}\) | \(=\) | \(\displaystyle v_2\) |

So $u_2$ and $v_2$ are adjacent if and only if $\map \phi {u_1}$ and $\map \phi {v_1}$ are adjacent.

Because $G_1$ is isomorphic to $G_2$, $\map \phi {u_1}$ and $\map \phi {v_1}$ are adjacent if and only if $u_1 = \map {\phi^{-1} } {u_2}$ and $v_1 = \map {\phi^{-1} } {v_2}$ are adjacent.

So $u_2$ and $v_2$ are adjacent if and only if $u_1 = \map {\phi^{-1} } {u_2}$ and $v_1 = \map {\phi^{-1} } {v_2}$ are adjacent.

That is, $G_2$ is isomorphic to $G_1$

Thus graph isomorphism preserves symmetry.

$\Box$

### Transitivity

Let $G_1$ be isomorphic to $G_2$ and $G_2$ be isomorphic to $G_3$.

Then there exist isomorphisms:

- $\alpha: \map V {G_1} \to \map V {G_2}$ from $\map V {G_1}$ to $\map V {G_2}$
- $\beta: \map V {G_2} \to \map V {G_3}$ from $\map V {G_2}$ to $\map V {G_3}$.

Consider the composite mapping $\beta \circ \alpha$.

From Composite of Bijections is Bijection, $\beta \circ \alpha$ is a bijection from $\map V {G_1}$ to $\map V {G_3}$.

Let $u_1, v_1 \in \map V {G_1}$.

Let $\map \alpha {u_1} = u_2$ and $\map \alpha {v_1} = v_2$.

Let $\map \beta {u_2} = u_3$ and $\map \beta {v_2} = v_3$.

We have that $\alpha$ and $\beta$ are isomorphisms.

So:

- $u_1$ and $v_1$ are adjacent if and only if $\map \alpha {u_1} = u_2$ and $\map \alpha {v_1} = v_2$ are adjacent
- $u_2$ and $v_2$ are adjacent if and only if $\map \beta {u_2} = u_3$ and $\map \beta {v_2} = v_3$ are adjacent.

Thus:

- $u_1$ and $v_1$ are adjacent if and only if $\map \beta {\map \alpha {u_1} } = u_3$ and $\map \beta {\map \alpha {v_1} } = v_3$ are adjacent.

That is, if and only if $\map {\paren {\beta \circ \alpha} } {u_1} = u_3$ and $\map {\paren {\beta \circ \alpha} } {v_1} = v_3$ are adjacent.

Thus it has been demonstrated that $\beta \circ \alpha$ is an isomorphism.

Thus graph isomorphism has been shown to preserve transitivity.

$\Box$

Graph isomorphism has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$