What are the epimorphisms in the category of Hausdorff spaces?

It appears to be the case that the epimorphisms in $\text{Haus}$ are precisely the maps with dense image. This is claimed in various places, but a comment on my blog has made me doubt the source I got my proof from (Borceux).

Borceux's argument crucially uses the following result:

If $A \subset X$ is a closed subspace of a Hausdorff space $X$, then the quotient $X/A$ is Hausdorff.

This appears to be false. As far as I can tell, if $X/A$ is Hausdorff, then $A$ and points in $X$ not in $A$ must be separated by open neighborhoods in $X$. But if this is true for every closed subspace $A$ of $X$, then $X$ is necessarily regular, and there are examples of Hausdorff spaces that aren't regular.

So: is it still true that the epimorphisms are precisely the maps with dense image? If so, what is a correct proof of this?

• You can also verify this by definition, considering the fact that you can factorize an arrow in an epi-mono in your category. The only other fact needed is that the equalizers in HTop are the closed subspaces. – Andy Oct 15 '12 at 10:33
• – commenter Oct 15 '12 at 11:02

Yes, according to Herrlich & Strecker, Section 6.10(4). Here’s the argument:

If $A\overset{f}\longrightarrow B$ is an epimorphism, let $C$ be the disjoint topological union of two ‘copies’ of $B$ where the corresponding points of the closure of $f[A]$ have been identified, and let $h$ and $k$ be the two natural maps from $B$ to $C$.

That’s as far as they actually write it out, but clearly the rest is that $h\circ f=k\circ f$, and $f$ is an epimorphism, so $h=k$, and $f[A]$ must therefore be dense in $B$.

Added: To see that $C$ is actually Hausdorff, let the copies of $B$ be $B_0=B\times\{0\}$ and $B_1=B\times\{1\}$, let $K=\operatorname{cl}f[A]$, and let $K_i=K\times\{i\}$ for $i\in\{0,1\}$. Finally, let $q:B_0\sqcup B_1\to C$ be the quotient map. Clearly $q(\langle x,i\rangle)$ and $q(\langle y,j\rangle)$ can be separated by disjoint open sets in $C$ whenever $x\ne y$, irrespective of $i$ and $j$. If $q(\langle x,0\rangle)\ne q(\langle x,1\rangle)$, then $x\in B\setminus K$, an open subset of $B$, so $q[B_0\setminus K_0]$ and $q[B_1\setminus K_1]$ are disjoint open nbhds of $q(\langle x,0\rangle)$ and $q(\langle x,1\rangle)$.

• Is it clear that $C$ is Hausdorff? – Qiaochu Yuan Oct 15 '12 at 3:05
• @Qiaochu: Yes. The only possible problem would be points $x$ and $y$ corresponding to the same point of $B$ but in different copies, and they have disjoint nbhds because we identified points of the closure of $f[A]$. Each of them has an open nbhd that misses the closure of $f[A]$ and so lives entirely in that point’s copy. – Brian M. Scott Oct 15 '12 at 3:08
• Great! So you also agree that Borceux's argument doesn't work? (Borceux's argument involves just quotienting $B$ by the closure of $f(A)$.) – Qiaochu Yuan Oct 15 '12 at 3:18
• @QiaochuYuan: I think a counterexample to Borceux's claim can be obtained by considering the topology $\tau$ on $\mathbb{R}$ generated by the set of rational numbers $\mathbb{Q}$ and the usual open intervals in $\mathbb{R}$. Since $\tau$ is finer than the usual topology, it is Hausdorff, but it is not regular: The set of irrational numbers $A = \mathbb{R} \setminus \mathbb{Q}$ is closed in $(\mathbb{R},\tau)$, but $1$ and $A$ do not have disjoint neighborhoods. – commenter Oct 15 '12 at 11:21
• I wouldn't give up on Borceaux's argument yet: it only needs that if $f:A\to B$ is an epimorphism in Haus, then $B/\overline{\mathrm{im}(A)}$ is Hausdorff, and this is certainly true, since we know it must be a one point space! :) (So the question is whether there is a proof of this short enough to qualify as a fix to Borceaux's argument rather than a different proof.) – Omar Antolín-Camarena Oct 16 '12 at 3:26

This should be a comment rather than an answer, but I don't have enough rep.

HTop is actually the largest subcategory of Top closed under finite limits (as computed in Top) where all maps with dense image are epi:

If $X$ is not Hausdorff then the equalizer of the projections $\pi_{1},\pi_{2}:X\times X\rightarrow X$, which is just the diagonal $\delta:X\rightarrow X\times X$, is not closed. (Recall that a space is Hausdorff iff the diagonal is closed.)

Let $C$ denote the closure of the diagonal in $X\times X$, let $d$ denote the factorization of $\delta$ through $C$, and let $p_{1}$ and $p_{2}$ denote the restrictions of $\pi_{1}$ and $\pi_{2}$ to $C$ respectively. Then the image of $X$ is dense in $C$ and $p_{1}\circ d = p_{2} \circ d$, but $p_{1}\neq p_{2}$, so $d$ is not epi.

This shows the fact Andy mentions in the comments, that equalizers are closed subspaces in HTop, is essential.

In Topology and Groupoids, p. 128, it is proved that an adjunction space $B \; _f\sqcup X$ is Hausdorff if (a) $B$ and $X$ are Hausdorff, (b) each $x \in X \backslash A$ has a neighbourhood closed in $X$ and not meeting $A$, and (c) $A$ is a neighbourhood retract of $X$.

I do not know if these conditions can be weakened for the case of $X/A$.

I should say this is really a comment on the comments rather than an answer to the question!