gn.general topology - Functions separting points in Hausdorff spaces

This is really a comment on t3suji's answer, but it's too long to be a comment as such.

t3suji's answer is the canonical one in the following precise sense. Let $e: X to Y$ be a morphism in any category. It's an elementary exercise to show that the following conditions on $e$ are equivalent:

1. $e$ is an epimorphism




2. the square
   $$
   begin{array}{ccc}
   X &stackrel{e}{to} &Y \
   edownarrow & &downarrow 1_Y \
   Y &stackrel{1_Y}{to} &Y
   end{array}
   $$
   is a pushout




3. for some morphism $f: Y to Z$, the square
   $$
   begin{array}{ccc}
   X &stackrel{e}{to} &Y \
   edownarrow & &downarrow f \
   Y &stackrel{f}{to} &Z
   end{array}
   $$
   is a pushout.

I'll only use the equivalence 1 $iff$ 3 here. The other implications are just scene-setting.

Suppose we want to show that a particular morphism $e$ is not epi. Assuming that there are enough pushouts around, we can argue as follows. Form the pushout square
$$
begin{array}{ccc}
X &stackrel{e}{to} &Y \
edownarrow & &downarrow f \
Y &stackrel{g}{to} &Z.
end{array}
$$
If $f neq g$ then the implication 1 $Rightarrow$ 3 tells us that $e$ is not epi. Moreover, this strategy is bound to work, in the sense that if $f = g$ then the implication 3 $Rightarrow$ 1 tells us that $e$ is epi after all.

It only remains to see that this is indeed what t3suji did. In his/her situation, $e$ was the inclusion $X to Y$. He/she then took the coequalizer of the two obvious maps $X to Y + Y$ (where $+$ means coproduct, i.e. disjoint union). For elementary and totally general reasons, this is the same thing as taking the pushout just mentioned. The morphisms that t3suji called $iota_1$ and $iota_2$, I called $f$ and $g$. Finally, although t3suji's pushout is in the category of all topological spaces, he/she then verified that the space $Z$ is indeed Hausdorff, from which it follows that it's also a pushout in Hausdorff spaces.

So now you know, in principle, how to answer any question of the form "prove that such-and-such a morphism isn't epi".