Exercise: Construct an explicit inverse of $\phi$ in the following theorem
Theorem: Given a short exact sequence with a right split (this is, a $v: K \to G$ such that $\epsilon v = 1$), which determines an action of $K$ in $H$ given by an $\omega: K \to{\rm Aut}\, H$ defined as $\omega(k)(h)=v(k)hv(k^{-1})$. Then $\exists \phi: H \times_{\omega}K \to G$ such that the following diagram commutes:
(using 1 instead of 0 since none of these groups require to be abelian)
$\mu_1, \mu_2$ and $\epsilon_2$ being homomorphisms given by:
$\phi$ is given by $\phi(h,k)= \mu(h)v(k)$ and is proven to be a group isomorphism in a theorem that is previously stated in the book. I just have to find a $\phi':G \to H\times_{\omega}K$ that cancels $\phi$. I could only find that $$\epsilon(\mu(h)v(k)) = \epsilon(\mu(h))\epsilon(v(k))$$ Since $\epsilon v = 1$ and $\ker\epsilon = \text{Im} \mu$
$$\epsilon(\mu(h))\epsilon(v(k))=1*k=k$$ So I know $\phi'$ can be something like $\phi'(g)= (?, \epsilon(g))$, but I'm completely lost on how to find a similar way for h, and I'm stuck, any help is appreciated (also, if I'm wrong in any way I also appreciate it)