I was asked to prove the following proposition
4.14 Proposition Let: $f: G \rightarrow H$ be a homomorphisms between groups $G$ and $H$.
(a) if $U$ is a subgroup of $G$ then $f(U)$ is a subgroup of $H$.
(b) if $V$ is a subgroup of $H$ then $f^{-1}(V)$ is a subgroup of $G$
I already proved part (a) and I am uploading my proof later today for feedback. My problem is in part (b). We don't know anything about $f^{-}1$. I don't know if $f$ being homomorphic implies $f^{-1}$ is homormorphic. if that's true then why is it true?. The proof would be the same if $f^{-1}$ is homomorphic
Actually, if a group homomorphism $f:G\to H$ has an inverse function $f^{-1}$, then $f^{-1}$ is automatically a group homomorphism too. Given any two $h_1, h_2\in H$, we can consider the two elements $f^{-1}(h_1)$ and $f^{-1}(h_2)$ in the condition for $f$ being a homomorphism and get $$ f\bigl(f^{-1}(h_1)*f^{-1}(h_2)\bigr) = f(f^{-1}(h_1))*(f(f^{-1}(h_2)) = h_1 * h_2 $$ Taking $f^{-1}$ on both sides of this yields $$ f^{-1}(h_1)*f^{-1}(h_2) = f^{-1}(h_1*h_2) $$ so $f^{-1}$ is a homomorphism.
But this is not really relevant for your exercise, because here the notation $f^{-1}(V)$ does not mean to apply an inverse function, but merely to take a preimage. By definition, $$ f^{-1}(V) \text{ means } \{ g\in G \mid f(g)\in V \} $$