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 $Z$.
Best 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.