The existence of the product of a morphism from the quotient of an equivalence relation. In Topologies et Faisceaux by Demazure, Proposition 3.3.4. p.180 (or p.192 in the linked pdf), states the following:

Let $R$ be a universal effective relation in $X$. Let $f:X\rightarrow Z$ be a compatible morphism with $R$, hence factorizing through $g:X/R\rightarrow Z$. TFAE.
(i). $g$ is a monomorphism.
(ii). $R$ is the equivalence relation defined by $f$.

Then it proceeds to  prove that (ii) implies (i) by way of Proposition 2.5 (the converse is indeed obvious). However, to use that proposition, it is necessary that the fibre product $X/R\times_ZX/R$ exists.
So my question is:

Why does the product $X/R\times_ZX/R$ exist, or does it necessarily exist?

It seems that this should follow from the universal effective hypothesis, or the universal effective "epicity" of $g$, but I cannot deduce the existence of the product in question from the given hypotheses.
Any help or reference is appreciated.
P.S. My last question turned out to be answered by the linked, newer version, so this time I checked that this is not covered in the pdf either, unfortunately.
 A: A possible answer, after some tries, follows.
(Please forgive the ugly-looking commutative diagrams; I wish I can use tikz here.)
We wan to show that if $R$ is the relation defined by $f$, then $g$ is a monomorphism.
Consider the diagram:
$$
\begin{array}{ccccc}
&&&&A\\
&&&&\downdownarrows\\
R&\rightrightarrows&X&\xrightarrow{p}&X/R\\
&&&f\searrow&\ \downarrow g\\
&&&&Z
\end{array}
$$
with $A\rightrightarrows X/R\xrightarrow{g}Z$ equal.
By hypotheses, $R$ is a universal effective equivalence relation, so $p$ is a universal effective epimorphism, and hence we can form two fibre products:
$$(X\times_{X/R}A),\,(X\times_{X/R}A)'.$$
Now consider the diagram:
$$
\begin{array}{ccccc}
&&(X\times_{X/R}A)\times_A(X\times_{X/R}A)'&\rightarrow&(X\times_{X/R}A)'\\
&&\downarrow&&\downarrow\\
&&(X\times_{X/R}A)&\rightarrow&A\\
&&\downarrow&&\downdownarrows\\
R&\rightrightarrows&X&\xrightarrow{p}&X/R\\
&&&f\searrow&\ \downarrow g\\
&&&&Z
\end{array}
$$
where $(X\times_{X/R}A)\times_A(X\times_{X/R}A)'$ exists as $(X\times_{X/R}A)\rightarrow A$ is also a universal effective epimorphism.
Notice the two morphisms $(X\times_{X/R}A)\times_A(X\times_{X/R}A)'\rightrightarrows X\xrightarrow{f}Z$ are equal to
$$(X\times_{X/R}A)\times_A(X\times_{X/R}A)'\rightarrow A\rightrightarrows X/R\xrightarrow{g}Z,$$
hence equal by assumption.
As $R$ is defined by $f$, we find then that the morphisms $(X\times_{X/R}A)\times_A(X\times_{X/R}A)'\rightrightarrows X$ factor through $R$.
This shows that the morphisms
$$(X\times_{X/R}A)\times_A(X\times_{X/R}A)'\rightarrow A\rightrightarrows X/R$$ are equal.
Finally, as compositions of universal effective epimorphisms are universal effective epimorphisms, we conclude that indeed the morphisms $A\rightrightarrows X/R$ are equal. Therefore $g$ is a monomorphism.
This shows the fibre product $X/R\times_{Z}X/R\cong X/R$, thus we can apply the proposition to show that $g$ is a monomorphism ... just kidding.
