Pullback $p^*$ reflects isomorphisms implies $p$ is universal quotient map $\newcommand{\im}{\operatorname{im}}$I'm trying to understand the implication $(i) \implies (ii)$ in the first theorem of "How algebraic is the change-of-base functor by Janelidze and Tholen, and I should preface this with "I'm not a topologist".
The statement is

Let $p \colon E \to B$ be a morphism in $\newcommand{\Top}{\operatorname{Top}} \Top$.
If the functor $p^* = E \times_B - \colon \Top/B \to \Top/E$ reflects isomorphisms, then $p$ is a universal quotient map.

Here:

*

*a functor $F \colon C \to D$ reflects isomorphisms if $Ff$ being an isomorphism implies that $f$ was already an isomorphism.


*$\Top/X$ is of course the category of spaces over $X$, i.e. tuples $(A, \alpha \colon A \to X)$ with morphisms $f \colon (A, \alpha) \to (C, \gamma)$ being morphisms of the underlying topological spaces satisfying $\gamma \circ f = \alpha$


*as the title implies, $E \times_B A$ for $(A, \alpha) \in \Top/B$ is the pullback of $E \xrightarrow{p} B \xleftarrow{\alpha} A$.


*universal quotient map basically means that first of all $p$ is a quotient map and that in any pullback $E \times_Z A$, the projection $\pi_2 \colon E \times_B A \to A$ is a quotient map.
Now, I practically don't understand most steps in the proof.
Here it is:


*

*Let $A'$ be the set $\im \pi_2$ equipped with the quotient topology for $\pi_2$.

*The inclusion $j \colon A' \to A$ is in $\Top/B$.

*$E \times_B j$ is a homeomorphism, thus $j$ is a homeomorphism.

*Thus $\pi_2$ is a quotient map.


My problems/troubles with this are:

*

*The general proof strategy seems to be to define a "quotient-like" map, and show that it is onto. But for that, they seem to be leaving out an assumption, right? Namely, that $p$ be a quotient map.  In particular, it's clear that if $p$ is not surjective, the map $\pi_2$ can only be onto if the image of $A \to B$ is contained in $\im p$.

So let's assume that $p$ is at least onto.

*

*In 2, why is $j \in \Top$? I'm not a topologist, and this is not trivial for me.
I understand that for $U \in A$ open, I have to check that $\{(e,a) | a \in U\} \subset E \times_B A$ is open, but I don't get it.

*How to see that $E\times_B j$ is a homeomorphism?


EDIT:
Ok, so assuming that $j$ is continuous, I now understand why $E(j) = E \times_B j$ is a homeomorphism.
Let $(A, \alpha)$ in $\Top/B$.
We can consider the (set-theoretical) corestricted mapping $\pi_2' \colon E \times_B A \to A' = \im \pi_2$ (i.e. we simply restrict the target of $\pi_2$ to the image). Then we consider $A'$ with the final topology for $\pi_2'$, so it is by construction a quotient map.
Denote again the inclusion by $j \colon A' \to A$; if $j$ is continuous, then it is clear that also $(A', \alpha' = \alpha j) \in \Top/B$.
Note also that $\pi_2 = j \pi_2'$.
We claim that
$\require{AMScd}$
\begin{CD}
E \times_B A @>{\pi_2'}>> A' \\
@V{\pi_1}VV @VV{\alpha'}V \\
E @>>{p}> B
\end{CD}
is a pullback.
So assume we have a span $E \xleftarrow{f} Q \xrightarrow{g}$ with $pf = \alpha' g$.
In other words, we have
\begin{align}
   p j = \alpha' g = \alpha (j g)
\ ,
\end{align}
thus there is a unique $u \colon Q \to E \times_B A$ such that
$$
  \pi_1 u = f \text{  and  } \pi_2 u = j g \ .
$$
But then $j \pi_2' u = j g$, and since $j$ is monic, $\pi_2' u = g$.
That is, $u$ satisfies
$$
  \pi_1 u = f \text{  and  } \pi_2' u = g \ ,
$$
so that the diagram is indeed a pullback diagram.
Thus $E \times_B A \cong E \times_B A'$.
In particular, $E \times_B j$ is an isomorphism, and so $j$ is an isomorphism, i.e.\ $A \cong A'$ and thus $\pi_2$ is a quotient map.

EDIT 2:
Continuity of $j$ is actually pretty easy, I think, at least if the following is correct.
Let $\pi_2'$ be as in the previous edit, i.e. the corestricted $\text{Set}$-map $\pi_2' = \pi_2 \colon E \times_B A \to A' = \im(\pi_2)$, and note that crucially $j \pi_2' = \pi_2$.
Also, $A'$ has final topology for $\pi_2'$.
But then
\begin{align*}
U \subset A \text{ open}
&\overset{\pi_2 \text{ cont.}}{\implies} \pi_2^{-1}(U) \text{ open}
\\
&\overset{(*)}{\iff} \pi_2^{-1}(U) = \pi_2'^{-1} \left( j^{-1} (U) \right) 
\text{ open}
\\
&\overset{\pi_2' \text{ quotient}}{\iff} j^{-1}(U) \text{ open}
\end{align*}
where $(*)$ is just the fact that taking preimages is a contravariant operation w.r.t. function composition.
 A: First, observe that the class of morphisms $p : E \to B$ such that $p^* : \textbf{Top}_{/ B} \to \textbf{Top}_{/ E}$ is conservative (= reflects isomorphisms) is closed under pullback: this is essentially an application of the pullback pasting lemma.
Thus, if we wish to show that every morphism in this class is a universal quotient map, it suffices to show that every morphism in this class is a quotient map.
Let $B'$ be the image of $p : E \to B$, considered as a subspace of $B$, and consider the pullback of the inclusion $B' \hookrightarrow B$.
The pullback of a subspace inclusion is a subspace inclusion, and since $B'$ is the image of $p : E \to B$, the pullback of the inclusion must be $\textrm{id} : E \to E$ (up to isomorphism).
But $p^* : \textbf{Top}_{/ B} \to \textbf{Top}_{/ E}$ is conservative, so that implies $B' \hookrightarrow B$ is also an isomorphism, hence $p : E \to B$ is surjective.
(I am heavily abusing notation here.
Really I am speaking of the inclusion $B' \hookrightarrow B$ considered as a morphism from $B' \hookrightarrow B$, considered as an object in $\textbf{Top}_{/ B}$, to $\textrm{id}_B : B \to B$, which is a terminal object in $\textbf{Top}_{/ B}$, etc.)
We can play the same game with the image of $p : E \to B$ topologised as a quotient $\tilde{E}$ of $E$ instead of as a subspace of $B$.
The continuous map $p : E \to B$ factors as a quotient map $E \twoheadrightarrow \tilde{E}$ followed by an injective continuous map $\tilde{E} \to B$.
The pullback of $\tilde{E} \to B$ is a bijective continuous map $E \times_B \tilde{E} \to E$.
But in fact we have a continuous inverse map $E \to E \times_B \tilde{E}$, namely the map given by the universal property of pullbacks whose first component is $\textrm{id} : E \to E$ and whose second component is the quotient map $E \twoheadrightarrow \tilde{E}$.
Thus $E \times_B \tilde{E} \to E$ is an isomorphism, so $\tilde{E} \to B$ is also an isomorphism.
Hence $p : E \to B$ is indeed a quotient map.
(The second paragraph above is actually redundant – but I include it for pedagogical purposes.)
