Hi there i need help with verifying if my proof is sound:

Let $f:X \rightarrow Y$ be a morphism of the category of Hausdorff topological spaces. Prove that $f$ is a epimorphism iff $f$ is a continuous map whose image is dense.

Proof: Recall that to show what morphisms are epimorphims we need to show for what conditions for $f:X \rightarrow Y$ we have for any parallel morphisms $h_1,h_2:Y\rightarrow Z$ with $h_1\circ f=h_2\circ f$ implies that $h_1=h_2$.

$"\Rightarrow"$: Let $f:X \rightarrow Y$ be a morphism where $X,Y \in \text{Ob}(\text{HausTop} )$. By way of contradiction assume that $f$ is not continuous map whose image is dense, i.e. $Y\not = \overline{\text{Im}f}$. Now we construct two parallel morphisms $h_1,h_2:Y\rightarrow Z$ with $Z=\{0,1\}$ equipped with the discrete topology and defined by $$h_1(y)=0 \,\,\, \forall y\in Y, \qquad h_2(y)= \begin{cases} 0 & y\in \overline{\text{Im }f} \\ 1 & y\not\in \overline{\text{Im }f} \end{cases}$$ Then we have $h_1\circ f=h_2\circ f$ but $h_1\not = h_2$, which contradicts the definition of epimorphism. Therefore $Y = \overline{\text{Im }f}$ must be true.

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    $\begingroup$ Why is $h_2$ continuous? $\endgroup$ – Thomas Andrews Mar 1 '18 at 21:51
  • $\begingroup$ Ahh dang, right I forgot to verify if $h_2$ was continuous, which obviously is not. Do you have any suggestions on what function might work to complete the contradiction? $\endgroup$ – Rami2314 Mar 1 '18 at 22:03
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    $\begingroup$ If $C$ is a closed subspace of a Hausdorff space $Y$, is $Y/C$ Hausdorff? If so, then you get two maps $Y\to Y/C$, one the constant map and the other the natural map. When $C=\overline{\operatorname{Im} f}$, $f$ equalizes these two functions. But I'm not sure $Y/C$ is always Hausdorff. $\endgroup$ – Thomas Andrews Mar 1 '18 at 22:10
  • $\begingroup$ Well, for that, I guess, we would need regularity ($T_3$) condition as well. $\endgroup$ – Berci Mar 1 '18 at 23:30
  • $\begingroup$ Related : math.stackexchange.com/questions/214045/… $\endgroup$ – Arnaud D. Mar 2 '18 at 11:16

This doesn't work because $h_2$ will usually not be continuous: $h_2^{-1}(\{0\})$ is probably not open.

Instead, you will have to do some real work to construct the space $Z$ you want to use. Hidden below is some discussion about how to think about a question like this.

In a category with pushouts, there is a "universal" way to test whether a map $f:X\to Y$ is an epimorphism, called the "cokernel pair". Namely, let $$\require{AMScd} \begin{CD} X @>{f}>> Y\\@V{f}VV @V{i}VV \\ Y @>{j}>> Z \end{CD}$$ be a pushout square. If $f$ is an epimorphism, then since $if=jf$ we have $i=j$. Conversely, if $i=j$, then $f$ is an epimorphism: given any object $A$ and maps $g,h:Y\to A$ with $gf=hf$, there is a unique map $p:Z\to A$ such that $g=pi$ and $h=pj$, and so $g=h$.

So to test whether $f$ is an epimorphism, we can form the pushout $Z$ of two copies of $f$ and ask whether the two inclusions $i,j:Y\to Z$ are equal. If you find pushouts in the category of Hausdorff spaces difficult to think about, you can just use the intuition behind them to explicitly construct a space $Z$ that works. The idea is that you are taking two copies of $Y$, but "gluing them together" along $f$ so that the two inclusions $Y\to Z$ are equal when composed with $f$. Think about how you might use the assumption that the image of $f$ is not dense to perform such a gluing and get a Hausdorff result.

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