$q:A\to B$ is a quotient map then $B$ homeomorphic to $A/\sim$ I was reading that if $q:(A,\mathcal{T}_A)\to (B,\mathcal{T}_B)$ is a quotient map then $B$ can be seen as a a quotient of $A$. Do we have $B$ homeomorphic to $A/\sim$ where $x \sim y \iff q(x)=q(y)$ (where $B$ is endowed with its topology and $A/\sim$ with the quotient topology)? How do you prove this? With the universal property of quotient?
 A: Let $\nu:A\to A/\sim$ denote the function prescribed by $x\mapsto[x]_{\sim}$. Then $\nu$ and $q$ are both quotient maps.
Function $\nu:A\to A/{\sim}$ respects function $q$ in the sense that
$q\left(x\right)=q\left(y\right)\implies\nu\left(x\right)=\nu\left(y\right)$.
This means exactly that $\nu=f\circ q$ for a function $f:A/{\sim}\to B$.
Since $q$ is a quotient map the continuity of $f\circ q$ implies
continuity of $f$.
Function $q:A\to B$ respects function $\nu$ in the sense that $\nu\left(x\right)=\nu\left(y\right)\implies q\left(x\right)=q\left(y\right)$.
This means exactly that $q=g\circ\nu$ for a function $g:B\to A/{\sim}$.
Since $\nu$ is a quotient map the continuity of $g\circ \nu$ implies
continuity of $g$.
Then $g\circ f\circ q=q$ and the surjectivity of $q$ makes us conclude
that $g\circ f$ is the identity on $A/{\sim}$. 
Similarly we conclude
that $f\circ g$ is the identity on $B$. 
Proved is now that $f$
and $g$ are homeomorphisms.
A: Let $\pi:A\to A/{\sim}$ be the projection map, that is, $\pi(x)=[x]$ for every $x\in A$, where $[x]=\{x'\in A\mid q(x)=q(x')\}$.
Since $q$ is a continuous function which is constant on each equivalence class of $A/{\sim}$, by universal property of quotient map,there exists a unique continuous function $\tilde q: A/{\sim} \,\to B$ such that the following diagram commutes: 
$$\require{AMScd}
\def\diaguparrow#1{\smash{
  \raise.6em\rlap{\scriptstyle #1}
  \lower.3em{\mathord{\diagup}}
  \raise.52em{\!\mathord{\nearrow}}
}}
\begin{CD}
&& A/{\sim}\\
& \diaguparrow{\pi} @VV \\\tilde q V \\
A @>> q> B
\end{CD},$$ that is, $q=\tilde q\circ \pi$.
Claim : $\tilde q $ is a homeomorphism.
Surjectivity : Let $y\in B$. Then there exists $x\in A$ such that $q(x)=y$(a quotient map, by definition, is surjective). Since the diagram commutes, we have $y=q(x)=\tilde q\circ\pi(x)=\tilde q([x])$, that is, $\tilde q$ is surjective.
Injectivity : Suppose $\tilde q([x])=\tilde q([y])$. Again by the commutativity of the diagram, it is easy to see that $q(x)=q(y)$. So $[x]=[y]$ by definition.
One last thing to check is the continuity of the inverse of $\tilde q$. For that, it is enough to prove that $\tilde q$ is an open map(why?).
Let $U$ be open in $A/{\sim}$. We need to prove that $\tilde q(U):=V$ is open in $B$. Since $q$ is a quotient map, $V$ is open in $B$ iff $q^{-1}(V)$ is open in $A$. Note that $$q^{-1}(V)=(\tilde q\circ \pi)^{-1}(V)=\pi^{-1}\circ \tilde q^{-1}(V)=\pi^{-1}\circ \tilde q^{-1}(\tilde q(U))=\pi^{-1}(U).$$ Remember that $U$ is open in $A/{\sim}$ iff $\pi^{-1}(U)$ is open in $A$. So $V$ is open in $B$.
Hence $\tilde q$ is a homeomorphism.   
