# Classifying the epimorphisms in the category of normal topological spaces and continuous maps

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 $$Z$$.

• "so the epimorphisms must be at least surjective" is backwards--rather, the conclusion is that at least all surjective maps are epimorphisms. – Eric Wofsey Sep 20 '19 at 19:18
• Does normal include Hausdorff? – Paul Frost Sep 20 '19 at 22:35
• @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. – Paul Frost Sep 21 '19 at 11:22
• @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. – Paul Frost Sep 21 '19 at 22:50
• @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. – Paul Frost Sep 21 '19 at 23:09

## 1 Answer

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.