Pushout in $\mathsf{Set}$ where one of the maps is injective From I.M. James' book General Topology and Homotopy Theory:

Suppose we have a cotriad $$X \xleftarrow{\xi}W \xrightarrow{\eta}
 Y.$$ ... we might expect the pushout of the cotriad to be a quotient
set of $X \sqcup Y$. So consider the triad $$X \xrightarrow{\sigma} X
 \sqcup Y \xleftarrow{\tau} Y.$$ The obvious relation $\sim$ on $X
 \sqcup Y$ to try is the one where $\sigma x \sim \tau y$ if there
exists a point $z \in W$ such that $\xi z = x$ and $\eta z = y$.
Unfortunately, this is not, in general, an equivalence relation... in general, it is necessary to form the transitive closure in order to obtain a quotient of $X \sqcup Y$ which satisfies the conditions for a push-out.
The procedure of forming the transitive closure can, however, be omitted if one of the insertions [EA: i.e. either $\xi$ or $\eta$] is
injective.

I just have no idea why the sentence in bold would be true. I can think of a situation where $\xi$ is injective, $\eta$ is not, and the relation is not transitive: suppose $\xi(b_1) = x_1, \, \xi(b_2) = x_2, \, \eta(b_1)= \eta(b_2) = y$. Then $x_1 \sim y$ and $x_2 \sim y$ but $x_1 \not \sim x_2$.
Normally I would think it was a mistake in the text, but I'm hoping I'm not missing something important about the pushout.
 A: If $X\stackrel f\leftarrow A\stackrel g\to Y$ is a cotriad, the pushout is
$$X\stackrel{\bar g}\to Z\stackrel{\bar f}\leftarrow Y$$
where $Z$ is the quotient space of $X\sqcup Y$ by the relation generated by $f(a)\sim g(a)$, and $\bar f$ is the composite $Y\to X\sqcup Y\to Z$. As often with relations, one only specifies what generates it, and the whole relation is the smallest equivalence relation containing these pairs. To obtain it, you can first take the reflexive and symmetric closure (but this is often implicitly assumed), then take the transitive closure. In the case of a pushout, one could also try to describe the relation directly by identifying
$$
f(a)\sim g(a)\\
f(a)\sim f(b)\text{ if $a,b$ have the same image under $g$} \\
g(a)\sim g(b) \text{ if $a,b$ have the same image under $f$}
$$
If one assumes reflexivity and symmetry, this may still not be transitive. But it is if $f$ is injective. So we don't have to take the transitive closure in that case. However, this is not what the author wrote. Maybe they had that in mind, but forgot to write it. So, no, you didn't miss anything about the pushout.
