Epimorphisms in TOP

Question

I have read in several places (e.g. in the wikipedia page and in this MSE question) that the epimorphisms in the category TOP of topological spaces with continuous maps are exactly the surjective continuous maps. However, I thought of an example that would contradict this statement. If I am wrong, could anybody help me figure out where?

Let $X=(0,1)$ and $Y=(0,1]$ with the subspace topology from $\mathbb{R}$. I claim that the inclusion $i:X\hookrightarrow Y$ is an epimorphism.

To prove that, consider two continuous functions $f,g:Y\to Z$, where $Z$ is any topological space, such that $f\circ i=g\circ i$. The goal is to prove that $f=g$. If $y\in(0,1)$, then $$f(y)=f\circ i(y)=g\circ i(y)=g(y)$$ and we are done. If $y=1$, then using that $f$ and $g$ are sequentially continuous $$f(1)=\lim_{n\to\infty}f\left(1-\frac{1}{n}\right)=\lim_{n\to\infty}f\circ i\left(1-\frac{1}{n}\right)=$$ $$\lim_{n\to\infty}g\circ i\left(1-\frac{1}{n}\right)=\lim_{n\to\infty}g\left(1-\frac{1}{n}\right)=g(1).$$

Hence, $f(y)=g(y)$ for all $y\in Y=(0,1]$, therefore $i$ is epimorphism which is not surjective.

Your argument works when $Z$ is Hausdorff but not necessarily otherwise — Jakobian, Apr 15 '25 at 09:47
The answer I linked explains what are epimorphisms in categories of topological and Hausdorff topological spaces. For topological spaces they are surjective continuous functions, for Hausdorff topological spaces they are continuous functions with dense images. — Jakobian, Apr 15 '25 at 09:52

Paul Frost · Accepted Answer · 2025-04-15T12:18:26.277

Your proof does not work. What you need here is that convergent sequences in $Z$ have unique limits. This property is known as the separation axiom US. Hausdorff implies US and US implies $T_1$.

As an example where your proof breaks down take $Z = (0,1]$ with the trivial topology. Define $f : Y \to Z, f(y) = y$, and $g : Y \to Z, g(y) = y$ for $y \in (0,1)$ and $g(1) = 1/2$. Both maps are continuous and $f \circ i = g \circ i$.

What you can prove is this:

Let $\text{Haus}$ be the full subcategory of Hausdorff spaces. The epimorphisms in $\text{Haus}$ are the continuous maps with dense image.

The proof is easy. If you have two maps $f, g : Y \to Z$ into a Hausdorff $Z$, then the set $A = \{y \in Y \mid f(y) = g(y) \}$ is closed in $Y$. Hence if $f$ and $g$ agree on a dense subset of $Y$, then they must agree on $Y$.

Epimorphisms in TOP

1 Answers1