Question: classify all epimorphisms in the full category $\mathcal{N}$ of Top on the normal topological spaces. Hint: you may find it useful to invoke Urysohn's lemma.

My work: I have no hypothesis so I'll just go of the definitions and see if something pops up. A morphism $f$ is epi if $hf = gf \implies h = g$ for all $g, h$. Let us say $X \xrightarrow{f} Y \xrightarrow{g, h} Z$. An epimorphism in Set is in particular an epimorphism in $\mathcal{N}$, so the surjective maps must be epimorphisms.

Suppose $g$ and $h$ are not equal. Then there is some $y \in Y$ such that $g(y) \neq h(y)$.

I'm not sure how to apply Urysohn's lemma, as the preimages of a point in $z$ under $g$ and $h$ aren't necessarily disjoint and those are the only non-trivial guaranteed closed sets in $Z$.

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    $\begingroup$ "so the epimorphisms must be at least surjective" is backwards--rather, the conclusion is that at least all surjective maps are epimorphisms. $\endgroup$ – Eric Wofsey Sep 20 '19 at 19:18
  • $\begingroup$ Does normal include Hausdorff? $\endgroup$ – Paul Frost Sep 20 '19 at 22:35
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    $\begingroup$ @WlodAA That is why I asked. Some authors define "normal" as in your comment, other authors additionally require $T_1$. Engelking is one of the latter ;-) See p. 62 of his great book. $\endgroup$ – Paul Frost Sep 21 '19 at 11:22
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    $\begingroup$ @WlodAA The latest edition seems to be from 1989. See heldermann.de/SSPM/SSPM06/sspm06.htm. Probably Engelking changed the definition between 1977 and 1989. And 1989 was a very good year for Poland and the world ... and for Germany on November 9, 1989. $\endgroup$ – Paul Frost Sep 21 '19 at 22:50
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    $\begingroup$ @WlodAA It wasn't easy at that time to gain access to literature published in the former Socialist Bloc. However, there was bookstore "Buchhandlung Harri Deutsch " whose owner must be have had special relations and offered a great variety of "inaccessible" books. Good luck for me - and good luck for everybody that those days are over. $\endgroup$ – Paul Frost Sep 21 '19 at 23:09

I once read somewhere that the epimorphisms in the category of Hausdorff spaces are precisely those continuous maps with dense image. So as a conjecture I began with that and here’s a solution (it turns out the epimorphisms are precisely the continuous maps with dense image by the way): Suppose $f$ is epi but $f$ does not have dense image. Then there is some $y \in Y$ such that $y \notin \overline{f(X)}$. Urysohn’s lemma now gives that there exists a continuous map $g : Y \mapsto [0,1]$ such that $g(y)=0$ and $g(z)=1$ for all $z \in \overline{f(X)}$. In particular $g(f(x))=1$ for all $x \in X$. So $$g \circ f =1 \circ f$$ where $1$ is the constant $1$ function. So $g=1$, so $0=g(y)=1$, contradiction. We conclude the image of $f$ lies dense in $Y$.

Suppose now $f$ has dense image. Now if $g_1,g_2: Y \rightarrow Z$ are such that they are equal on $f(X)$, $g_1$ and $g_2$ are two continuous maps that are equal on a dense subset, so if we’re lucky they are equal ;). So if we prove the following we’re done:

Lemma Let $h_1,h_2: A \rightarrow B$ be continuous maps between Hausdorff topological spaces that are equal on a dense subset. Then $h_1=h_2$.

Proof We will show the set $C=\{ x \in A | h_1(x) \neq h_2(x) \}$ is open in $A$, from this the result immediately follows. Since $A$ is Hausdorff, the diagonal $\Delta \subset A \times A$ is closed. So the inverse image of $\{(x,y) | x \neq y\}$ under the composition $$A \xrightarrow{x \mapsto (x,x)} A \times A \xrightarrow{(h_1,h_2)} A \times A$$ is open. But this inverse image is precisely $C$ and we’re done.


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