$f:X \to Y$ is quotient map if $X/f \cong Y$

Question

It seems to me that in the answer of When is a space homeomorphic to a quotient space? only one direction has been adressed.

So let $f:X \to Y$ be a surjective continuous map and let $h:X/f \to Y$ be a homeomorphism. I don't quite see why $f$ should be a quotient map.

Here, by quotient map I mean a surjective continuous map $q:X \to Y$, such that whenever $q^{-1}(A) \subseteq X$ is open, $A \subseteq Y$ is open.

By the universal property of $\pi_f:X \to X/f$ we have continuous bijections $\bar{f}:X/f \to Y$ and $\overline{h^{-1}\circ f}:X/f \to X/f$ but I can't show that these are open. Another attempt was to show that $h^{-1}\circ f$ has the same kind of universal property as $\pi_f$, but I didn't manage..

Answer 1

The formulation should be more precise: if $f: X \to Y$ is surjective and continuous define $R_f$ on $X$ by $x R_f x' \iff f(x)=f(y)$, the equivalence relation induced by $f$, and let the canonical quotient map be denoted $q_f: X \to X{/}f$, sending $x$ to its class $[x]$ under $R_f$. The function $\bar{f}: X{/}f \to Y$ (defined by $\bar{f}([x])=f(x)$ obviously ) is then well-defined set-theoretically and a bijection.

The fact the OP refers to:

$f$ is a quotient map iff $\bar{f}: X{/}f \to Y$ is a homeomorphism.

So no arbitary $h$ is involved at all.

The left to right implication is quite easy: $\bar{f}$ is continuous as $\bar{f} \circ q_f = f$ and $f$ is continuous (using the universal property for the quotient topology induced by $q_f$) and if $O \subseteq X{/}f$ is open then $\bar{f}[O] = f[O]$ which is open in $Y$ as $f^{-1}[f[O]] = q_f^{-1}[O]$ is open in $X$ and $f$ is assumed to be quotient, so that $\bar{f}$ is open and we already know from set-theory it's a bijection.

The reverse implication is quite similar. Only "$O \subseteq Y$ and $f^{-1}[O]$ open in $X$ implies $O$ open" needs to be shown. And this is clear as $O=\bar{f}[\bar{f}^{-1}[O]]$ and $\bar{f}$ is open.