Functions separting points in Hausdorff spaces - MathOverflow

Functions separting points in Hausdorff spaces

2010-02-01T14:57:31Z

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.~~

Answer by t3suji for Functions separting points in Hausdorff spaces

2010-02-01T17:54:48Z

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.

Answer by Tom Leinster for Functions separting points in Hausdorff spaces

2010-02-01T20:31:46Z

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:

$e$ is an epimorphism
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
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".