Composition of two functions, ${f_1}$ and ${f_2}$, is commutative if ${f_1} \circ {f_2} = {f_2} \circ {f_1}$. Even when the functions are bijective, it is not necessary that their composition is commutative. My question is, given two specific bijective functions, isomorphism and complementation on a graph $G$, is their composition commutative?
I realized that if two graphs $G$ and $H$ are isomorphic, then their complements are also isomorphic. This led me to see (ref below proof by contradiction) that the above composition is commutative. Can anyone pls tell me if this is correct! It would be great if you could point out any other theory that has led to similar conclusion.
My reasoning: By contradiction, ${f_1} \circ {f_2} (G) \not= {f_2} \circ {f_1} (G)$.
$=> {f_1} ({f_2}(G))={f_1} (G^{c}) \not= {f_2} ({f_1} (G))= {f_2} (H) = H^{c}$
However, $H^{c}$ and $G^{c}$ are isomorphic.
If you try to think of the operation of complementation as a function you are making things harder to understand than they need to be. You would have to carefully define a class of 'all' graphs. Whether it is an isomorphism on such a class is not relevant to your actual question as I understand it.
If $G$ and $H$ are finite simple graphs and $\varphi: G \rightarrow H$ is a graph isomorphism then $\varphi(G)^C = \varphi(G^C)$ because these two graphs have the vertices and the same edges. So you could say that complementation and applying the isomorphism commute.