Proving that $f^{-1} \in \operatorname{Hom}(Y,X).$

Question

I want to prove this question:

If $X,Y$ are groups and $f\in \operatorname{Hom}(X,Y)$ is bijective, then $f^{-1} \in \operatorname{Hom}(Y,X).$ Could anyone tell me how to start, please?

The point is that my professor defined $f\in \operatorname{Hom}(X,Y)$ to be an isomorphism if it is bijective and $f^{-1} \in \operatorname{Hom}(Y,X)$ and he said in groups it is enough to say bijective and to convince us with that he gave us the above problem to try to solve.

Answer 1

You have to show that $f^{-1}(xy)=f^{-1}(x)f^{-1}(y)$. Since $f$ is bijective, there exists unique $u,v\in X$ such that $f(u)=x,f(v)=y$, we deduce that $f^{-1}(x)=u, f^{-1}(v)=y$. Since $f$ is a morphism of group, $f(uv)=f(u)f(v)=xy$ implies that $f^{-1}(xy)=uv=f^{-1}(x)f^{-1}(y)$.

Answer 2

Since $f$ is bijective, you know $f^{-1}$ exists. Prove that $f^{-1}$ is a homomorphism.