# Functions separting points in Hausdorff spaces

A colleague in algebra asked me this, and I couldn't answer it. On the Wikipedia page for "epimorphism" it is claimed that in the category of Hausdorff spaces and continuous maps, a function is epi if and only if it has dense range. The "if" case is easy, but I couldn't justify the "only if" case.

This boils down to: let Y be a Hausdorff space, and let X in Y be a closed subset not equal to Y, and not empty. Can you find a Hausdorff space Z and functions f,g:Y->Z such that f and g agree on X, but are not equal. I think, by using a quotient argument, you can assume that X is just a point.

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I am not convinced you can reduce the general case to one where X is a point. You would need better separation than Hausdorff, or the quotient may not be Hausdorff. – Harald Hanche-Olsen Feb 1 '10 at 15:10
Sorry, yes, that's right: if you collapse X to a point x then it's not clear that you can separate x from other points. – Matthew Daws Feb 1 '10 at 15:15
If you can reduce to the case of a point, then you are done. Take Y=Z, f to be the identity and g to be the map sending all of Y to the point X. So all the challenge is in that reduction step. – David Speyer Feb 1 '10 at 16:27
@frb: Your second map is likely to be discontinuous. – Harald Hanche-Olsen Feb 1 '10 at 17:04
Answer withdrawn: sorry, missed the obvious fact that the map isn't open! – Charles Siegel Feb 1 '10 at 17:07

Let $Y$ be a Hausdorff space, and let $X\subset Y$ be a closed subspace. Consider disjoint union of two copies of $Y$, and let $Z$ be the coequalizer of two embeddings of $X$ into it (that is, we glue two copies of $Y$ along $X$). Clearly, the two natural maps $\iota_{1,2}:Y\to Z$ coincide only on $X$. It is easy to see that $Z$ is Hausdorff.

Indeed, take $z_1,z_2\in Z$, $z_1\ne z_2$. The map $p:Z\to Y$ is continuous, so if $p(z_1)\ne p(z_2)$, we can take preimages of open neighborhoods of $p(z_1)$ and $p(z_2)$ to separate $z_1$ and $z_2$. It remains to deal with the case $z_{1,2}=\iota_{1,2}(y)$ for $y\in Y-X$. But the neighborhoods $\iota_{1,2}(Y-X)$ work.

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

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

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Of course, in general the Principle of Conservation of the Difficulty under Changes in the Statement will probably kick in in most cases! – Mariano Suárez-Alvarez Feb 1 '10 at 20:37
Mariano, you mean that rephrasing a question in a different language never makes it fundamentally easier? I don't think I agree, and that wasn't the point of my answer. I was just observing that t3suji's answer (man, typing that name is giving me trouble) embodies some kind of universal method for solving these problems. – Tom Leinster Feb 1 '10 at 21:31
Yep: I think this is a very nice observation by Tom. Actually, I sat in on an algebra grad course last year which dealt with some of this homological algebra. This popped into my mind immediately upon seeing t3suji's answer. So I think I disagree with Mariano: if someone had said "pushout" to me, and walked away, I think I would have come up with t3suji's idea pretty quickly... – Matthew Daws Feb 1 '10 at 21:47
@Tom Leinster: (man, typing that name is giving me trouble) Sorry for that --- I somehow got so used to it that I have entered it in Math Overflow without much thought. It feels sort of silly to change it now... @Matthew Daws: It's interesting that somehow pushouts are not the first thought after reading the original problem (at least not for me). So maybe Tom's beautiful abstract set-up should help me see it next time I run across it. – t3suji Feb 1 '10 at 22:00