We can try to build an isomorphism. How about we say \(f(a) = b\text{,}\) \(f(b) = c\text{,}\) \(f(c) = d\) and \(f(d) = a\text{.}\) This is definitely a bijection, but to make sure that the function is an isomorphism, we must make sure it respects the edge relation. In \(G_1\text{,}\) vertices \(a\) and \(b\) are connected by an edge. In \(G_2\text{,}\) \(f(a) = b\) and \(f(b) = c\) are connected by an edge. So far, so good, but we must check the other three edges. The edge \(\{a,c\}\) in \(G_1\) corresponds to \(\{f(a), f(c)\} = \{b,d\}\text{,}\) but here we have a problem. There is no edge between \(b\) and \(d\) in \(G_2\text{.}\) Thus \(f\) is NOT an isomorphism.
in-context