Suppose $q\colon X\to X/\sim$ is a quotient map in the category of topological spaces $\mathsf{Top}$.
Why is $q$ an extremal epimorphism? I suppose there is a factorization $q=i\circ p$, with $i$ a monomorphism, hence injection. Since $q$ is surjective, so is $i$, and so $i$ is a continuous bijection. Why is it a homeomorphism? If $U\subseteq X/\sim$ is open, then $q^{-1}(U)$ is open, and $q(q^{-1}(U))=U$ since $q$ is surjective. Then
$$
i^{-1}(U)=i^{-1}(q(q^{-1}(U))=p(q^{-1}(U)).
$$
But I don't know if this is the right direction. Is $p$ necessarily open?
Best Answer
First note (or know, or prove) that in $\mathsf{Top}$ a monomorphism is a continuous injection (and an epimorphism is a continuous surjection), see Wikipedia e.g.
So write $q = m \circ f$ where $m:Y \to Z$ and $f: X \to Y$ is continuous and $q: X \to Z$ being the given quotient map (or take $Z=X{/}{\sim}$ if you like) and $m$ is an injective continuous map. We need to show that $m$ is a homeomorphism (i.e. an isomorphism in $\mathsf{Top}$).
So $m$ needs to be open and surjective as well. Openness is equivalent to the inverse being continuous (for bijections).
$m$ is surjective: if $z \in Z$ we find $x \in X$ with $q(x)=z$ (quotients maps are surjective) and then $m(f(x)) = q(x)= z$, showing that $m$ is surjective.
Openness of $m$: let $O$ be open in $Y$ and note that $m[O]$ is open in $Z$ iff $q^{-1}[m[O]]$ is open in $X$ (by definition of the quotient topology/quotient maps) and simple set theory tells us:
$$q^{-1}[m[O]] = (m \circ f)^{-1}[[m[O]] = f^{-1}[m^{-1}[m[O]] = f^{-1}[O]$$
where the last step uses injectivity of $m$. So $q^{-1}[m[O]]$ is open by continuity of $f$ and so $m[O]$ is open and we are done.