# Graph Theory: Isomorphic graphs

Show that the inverse of an isomorphism of graphs is also an isomorphism of graphs.

So I just started a graph theory course and am having a little trouble with one of the problems on the homework. I know that a graph is isomorphic if there are bijections Θ:V(G)→V(H) and Φ:E(G)→E(H) such that Ϯ_G(e)=uv if and only if Ϯ_H(Φ(e))=Θ(u)Θ(v). That is, they have the same structure, but differ only in the names of the vertices and edges. I just don't know how to find the inverse of an isomorphic graph. Any help would be great thanks.

• Hint: Since $\Theta$ and $\Phi$ are bijections, thus they have inverses $\Theta^{-1}$ and $\Phi^{-1}$. So you just have to show that $I_H(e) = uv$ iff $I_G(\Phi^{-1}(e)) = \Theta^{-1}(u)\Theta^{-1}(v).$ Also, you're not trying to find the inverse of an "isomorphic graph," so much as an inverse of a graph isomorphism. – goblin Jan 13 '14 at 21:19
• So since Θ and Φ are bijections, they have inverses Θ^(−1) and Φ^(−1). But how do I show that I_H(e)=uv iff I_G(Φ^(−1)(e))=Θ^(−1)(u)Θ^(−1)(v)? This is my first homework so if you could just explain it to me, I would really appreciate that. Thanks – Brian Jan 13 '14 at 22:05
• Sure, I just wrote up an answer. Its not especially well-written though, so I suggest you put it into your own (probably better) words before handing it in. – goblin Jan 13 '14 at 22:19
• Thank you, that helped a lot. – Brian Jan 13 '14 at 23:03

Since $\Theta$ and $\Phi$ are bijections, thus they have inverses $\Theta^{-1}$ and $\Phi^{-1}$.
Now let $e$ denote an edge of $H$ and $u,v$ denote vertexes. All that remains is to show the following equivalence. $$I_H(e)=uv \iff I_G(\Phi^{−1}e)=(\Theta^{-1}u)(\Theta^{-1}v).$$
1. $I_H(e)=uv$
2. $I_H(\Phi \Phi^{-1}e) = (\Theta\Theta^{-1}u)(\Theta\Theta^{-1}v)$
3. $I_G(\Phi^{-1}e) = (\Theta^{-1}u)(\Theta^{-1}v)$