Clearly, if a map between Stone spaces is surjective on points it is an epimorphism. In the category of topological spaces, surjections coincide with epimorphisms. In the category of Hausdorff spaces, epimorphisms are precisely the continuous functions with dense image: in one direction, dense maps are epis since equalizers are closed and $hf=kf$ iff $\mathrm{im}(f)$ is contained in the equalizer of $h$ and $k$; in the other direction, if $f:X\to Y$ doesn't have dense image, then it coequalizes the two distinct maps $i,j: X\to (X+X)/\sim$ where $\sim$ identifies the two copies of the closure of the image. I think if the construction of $(X+X)/\sim$ could be performed in the category of Stone spaces it would show that all epimorphisms in this category are surjective, since $\mathrm{im}(f)$ is always closed whenever $X$ is compact Hausdorff. However, I don't know if Stone spaces are closed under the necessary pushouts as a subcategory of topological spaces.
Are epimorphisms in the category of Stone spaces surjective
category-theorygeneral-topology
Related Solutions
Yes, according to Herrlich & Strecker, Section 6.10(4). Here’s the argument:
If $A\overset{f}\longrightarrow B$ is an epimorphism, let $C$ be the disjoint topological union of two ‘copies’ of $B$ where the corresponding points of the closure of $f[A]$ have been identified, and let $h$ and $k$ be the two natural maps from $B$ to $C$.
That’s as far as they actually write it out, but clearly the rest is that $h\circ f=k\circ f$, and $f$ is an epimorphism, so $h=k$, and $f[A]$ must therefore be dense in $B$.
Added: To see that $C$ is actually Hausdorff, let the copies of $B$ be $B_0=B\times\{0\}$ and $B_1=B\times\{1\}$, let $K=\operatorname{cl}f[A]$, and let $K_i=K\times\{i\}$ for $i\in\{0,1\}$. Finally, let $q:B_0\sqcup B_1\to C$ be the quotient map. Clearly $q(\langle x,i\rangle)$ and $q(\langle y,j\rangle)$ can be separated by disjoint open sets in $C$ whenever $x\ne y$, irrespective of $i$ and $j$. If $q(\langle x,0\rangle)\ne q(\langle x,1\rangle)$, then $x\in B\setminus K$, an open subset of $B$, so $q[B_0\setminus K_0]$ and $q[B_1\setminus K_1]$ are disjoint open nbhds of $q(\langle x,0\rangle)$ and $q(\langle x,1\rangle)$.
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.
Best Answer
For a direct topological proof, suppose $f:X\to Y$ is not surjective and let $y\in Y\setminus f(X)$. In a Stone space, points can be separated by clopen sets, and then by a compactness argument points and closed sets can be separated by clopen sets. So, there is a clopen set $C\subseteq Y$ such that $y\in C$ and $f(X)\subseteq Y\setminus C$. We now have $1_Cf=0f$ where $1_C,0:Y\to\{0,1\}$ are the characteristic function of $C$ and the constant $0$ map, respectively. This witnesses that $f$ is not epic in the category of Stone spaces.