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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?

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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
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3 Answers

up vote 12 down vote accepted

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 $q(\langle x,1\rangle)$.

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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
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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.

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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!

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