My apologies if this a naive question.
Let $\phi:G\rightarrow H$ be a continuous surjective homomorphism of topological groups. Denote the kernel of $\phi$ by $N$, then: $$\begin{align} f:G\times N&\longrightarrow G\\ (g,n)&\longmapsto g\cdot n \end{align}$$ is a continuous map. Indeed let $U\subset G$ be an open set, then if $\mu:G\times G$ denotes the multiplication map, we have that: $$f^{-1}(U)=\mu^{-1}(U)\cap (G\times N)$$ Since $\mu^{-1}(U)$ is open in $G\times N$, it follows that $f^{-1}(U)$ is open in $G\times N$, when equipped with subspace topology. Therefore $N$ acts continuously on $G$ from the right. We can take the topological quotient, given by the equivalence relation: $$g\sim h\Leftrightarrow \exists n\in N, g=h\cdot n$$ this is the same equivalence relationship defining $G/N$, and an isomorphism theorem guarantees that $G/N\cong H$, as groups.
We have by assumption that $\phi$ is continuous, and also satisfies all the properties of an algebraic quotient map. However, one can show that the topological quotient map $\pi:G\rightarrow G/N$ is always open even in the case where $G$ is just an arbitrary topological space and $N$ acts on it continuously from the right. So, what is stopping $\phi$ from being the topological group? I know that there is an open mapping theorem for topological groups which places considerable restrictions on the possible groups, so something is wrong with just saying $\phi$ is also the topological projection map, but I don't see why that is...