"Let $\phi : H \rightarrow Aut(N)$ be a group homomorphism (H and N are groups). Define two group homomorphisms: $\epsilon_N : N \rightarrow N \rtimes_\phi H$ which maps $n$ to $(n,1)$ and $\epsilon_H : H \rightarrow N \rtimes_\phi H$ which maps $h$ to $(1,h)$.
Let $f_N : N \rightarrow G$ and $f_H : H \rightarrow G$ be group homomorphisms that satisfy: $f_H(h)f_N(n)f_H(h)^{-1}=f_N(\psi_h(n))$
Show that there exists a unique group homomorphism $f : N \rtimes_\phi H \rightarrow G$ satisfying $f \circ \epsilon_N = f_N$ and $f \circ \epsilon_H = f_H$."
$\dots$
All my attempts to construct such a homomorphism have been in vain, I figured $f((n,h))$ would simply equal a product such as $f_N(n)f_H(g)$ or $f_N(\psi_h(n))f_H(h)$, but then I can't prove $f$ is a homomorphism if G is not abelian, so I can't even prove existance, let alone uniqueness. I tried constructing a picture of the mappings and that didn't help, so I'm led to believe I don't have to actually construct $f$, but there is some rules about semiproducts or something about the relation $f_N(\psi_h(n))$ holds that I am not picking up on.