In Topologies et Faisceaux by Demazure, during the second part of the proof of Proposition 1.8, one finds the statement that (background at the end) the morphism $U\times_ST\rightarrow U$ is an effective epimorphism, and the reason given is that because it has a section over $U$.
My Question:
Why is the morphism $U\times_ST\rightarrow U$ an effective epimorphism if it has a section?
I know why it has a section; it arises from the two $S$-morphisms $U\rightarrow U$ and $U\rightarrow T$. But I don't know why this implies that it is an effective epimorphism. While it is easy to show that it is epimorphic by the existence of a section, I found no clues as to how to show the effectivity. I think this should come from some simple arguments, but cannot figure it out.
Thanks in advance for help or references.
Some backgrounds:
We are consider $U\overset{v}\rightarrow T\overset{u}{\rightarrow}S$ where $uv$ is a universal effective epimorphism and $u$ is quarrable, which means the fibre product with every other morphism $S'\rightarrow S$ exists. And we want to show that $u$ is a universal effective epimorphism.
And a morphism $u:U\rightarrow V$ is an effective epimorphism if the diagram $$V\leftarrow U\leftleftarrows U\times_VU$$ is exact.
Suppose $e : A \to E$ has a section $e' : E\to A$. Then $e$ is the coequalizer of $\operatorname{id}_A$ and $e' \circ e$:
We have $e\circ \operatorname{id}_A = e\circ e' \circ e$: Let $q : A\to Q$ be such that $q\circ \operatorname{id}_A = q\circ e' \circ e$. If there exists an $u : E \to Q$ such that $u\circ e = q$ then it is unique since $u = q\circ e'$. Defining $u:= q\circ e'$ shows existence.
This shows that $e$ is a regular epi. Finally observe that a regular epi with a kernel pair is also the coequalizer of its kernel pair meaning it is effective.