We have a function $\varphi:G\rightarrow H$ is an isomorphism, show its inverse $\varphi^{-1}:H\rightarrow G$ is also an isomorphism
I am fine with showing it to be a homomorphism and surjective, this is my attempt at showing injectivity, I'm unsure if they would be sufficient
Assume: $\forall h\in H, \exists g\in G$ s.t. $\varphi(g)=h$ and $\forall g_1, g_2\in G, \varphi(g_1)=\varphi(g_2)\Rightarrow g_1=g_2$
Injectivity – Need to show $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=g_2$
$\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=\varphi(\varphi^{-1}(g_2))$ from our surjectivity assumption on $\varphi$ we know there exists such a $\varphi^{-1}(g_2)\in H$ and from our injectivity assumption on $\varphi$ we know that this solution is unique.
[Sufficient? What more could be added? How could I do this clearer?]
Best Answer
Your question has nothing to do with group theory; the fact that inverse functions are necessarily bijective is a matter of set theory. And if you know, as a suppose, that in order to have an inverse function, a function $f$ must be bijective, it is pretty obvious that the inverse $f^{-1}$ will always be bijective. After all the requirement for an inverse is that both $f^{-1}\circ f$ and $f\circ f^{-1}$ are identity functions of their respective domains; then from the symmetry of this requirement it is immediate that $f$ is also the inverse of $f^{-1}$. But then $f^{-1}$ always has an inverse function ($f)$, so it must be bijective.
One can of course deduce from this a more detailed argument for why $f^{-1}$ must be injective. Recall that the have an inverse $f$ must be injective, since if $f(x)=f(y)$ then applying the supposedly existing inverse $f^{-1}$ to both sides one finds $f^{-1}(f(x))=f^{-1}(f(y))$ which becomes $x=y$. Now switching roles of $f$ and $f^{-1}$, if one has $f^{-1}(x)=f^{-1}(y)$ then applying $f$ to both sides gives $x=y$.