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Let $A$ be a dense subset of a topological space $X$. Then the inclusion map $i:A\to X$ is an epimorphism.

Is this true? If so, how do I prove it? I know the set up. Suppose $j,h:X\to Y$ are continuous functions such that $ji=hi$. Suppose $x\in X$. Then we want to prove that $j(x)=h(x)$.

Where does being dense come in?

Jacob Laurinitus
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You need to limit the category a little more (e.g. to Hausdorff spaces) for this to be true.

If, for instance, we are allowed to equip $Y$ with the indiscrete topology, then $j,h$ can be any functions that agree on $A$, and it should be apparent that if $A\neq X$, then we could extend these functions arbitrarily.

That is, in the category of topological spaces, then $$\text{the inclusion } i : A \to X \text{ is an epimorphism } \iff A=X$$

If you work in the category of Hausdorff spaces, then the inclusion of a dense set into a space will always be an epimorphism. This likely is possible with less powerful separation axioms (e.g. $T_1$), but I don't know precisely how weak we can make them.

Brian Moehring
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    I don’t think you can weaken it much. See this question. For a space $Y$, the property that any two continuous $f,g\colon X\to Y$ that agree on a dense subset of $X$ must agree on all of $X$ is equivalent to $Y$ being Hausdorff. – Arturo Magidin Aug 17 '19 at 04:59