I want to prove the following version of the splitting lemma. Let $G,K$ be topological groups. Let $1\to G\xrightarrow{\phi}H\xrightarrow{\eta}K\to1$ be a central extension of $K$ by $G$. Then $\phi$ and $\eta$ are continuous homomorphisms and $\phi(G)\cong \ker(\eta)\subseteq Z(H)$. Suppose this extension is trivial, i.e. there is a section $\mu:K\to H$ such that $\eta\circ\mu=Id_K$. Then $H\simeq G\times K$.
I define $\Psi:G\times K\to H$ by $\Psi(g,k)=\phi(g)\mu(k)$. Continuity of this map is clear. Also, it is a homomorphism since $\phi(g)\in Z(H)$ for all $g\in G$. I want to show it is an isomorphism, but I am stuck proving both surjectivity and injectivity. To construct an inverse map $\Phi:H\to G\times K$, I want to do something like $\Phi(h)=(…,\eta(h))$ and using that $\mu$ is a section, but I can't figure it out.
Surjectivity
For $h\in H$ consider $x=h(\mu\circ\nu)(h^{-1})$.
This gives $\nu(x)=\nu(h)(\nu\circ\mu\circ\nu)(h^{-1})=\nu(h)\nu(h^{-1})=1_K$.
Therefore there is some $g\in G$ with $\phi(g)=x$.
Together we have $h=x(\mu\circ\nu)(h)=\Psi(g,\nu(h))$
Injectivity
If $\Psi(g,k)=1$ then $\phi(g)\mu(k)=1_H$ so $\nu\circ\mu(k)=1_K$ so $k=1_K$.
Therefore $\phi(g)=1_H$ so $g=1_G$.