Graph Isomorphism is Equivalence Relation/Proof 2
Theorem
Graph isomorphism is an equivalence relation.
Proof
In the following, let:
\(\ds G_1\) | \(=\) | \(\ds \struct {\map V {G_1}, \map E {G_1} }\) | ||||||||||||
\(\ds G_2\) | \(=\) | \(\ds \struct {\map V {G_2}, \map E {G_2} }\) | ||||||||||||
\(\ds G_3\) | \(=\) | \(\ds \struct {\map V {G_3}, \map E {G_3} }\) |
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:
\(\ds \map \phi {u_1}\) | \(=\) | \(\ds u_2\) | ||||||||||||
\(\ds \map \phi {v_1}\) | \(=\) | \(\ds v_2\) |
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.
$\blacksquare$
Sources
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs: Theorem $2.3$