7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ A concrete category over a category $\mathbf{X}$ is a pair of the form $(\mathbf{A},U)$ where $U:\mathbf{A}\to\mathbf{X}$ is a faithful functor. If in particular if $\mathbf{X}$ is $\mathbf{Set}$ and $U(f)$ is surjective, then $f$ is a epimorphism. $\endgroup$ – user170039 Apr 14 '20 at 5:49
8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Nice! If I'm not mistaken we could use a similar argument to show that epimorphisms in compact Hausdorff spaces are surjective, using the fact that such spaces are completely regular? $\endgroup$ – Jonas Frey Apr 14 '20 at 4:18
  • $\begingroup$ Yep, that would work. $\endgroup$ – Eric Wofsey Apr 14 '20 at 4:34
9
$\begingroup$

The category of Stone spaces is dually equivalent to that of Boolean algebras, so an epimorphism of Stone spaces is exactly the same as a monomorphism between the underlying algebras.

But now a monomorphism of Boolean algebras is injective (the forgetful functor to $\mathbf{Set}$ preserves monomorphisms, because it has a left adjoint); so we may ask : let $f: B\to B'$ be an injective morphism of Boolean algebras, and let $U\in S(B)$ be an ultrafilter. Is there an ultrafilter $U'$ on $B'$ such that $f^*U' = U$. But here $f^*U'$ is simply $f^{-1}U' = B\cap U'$ if we treat $B$ as a subalgebra of $B'$.

And then the answer is clearly yes : let $U$ be an ultrafilter on $B$. Then look at its image in $B'$ : it's a filter, and is therefore contained in some ultrafilter $U'$. Now $U'\cap B$ is an ultrafilter, which contains $U$ and must therefore be equal to $U$.

So $f^*$ is surjective, and any epimorphism is of the form $f^*$ for some monomorphism $f$ (up to isomorphism), so we are done.

$\endgroup$
1
  • $\begingroup$ Great! I accepted the more direct proof, but this one is also very helpful. so essentially you give a positive answer to the question whether Bool is an injective object in the category of Boolean algebras. $\endgroup$ – Jonas Frey Apr 13 '20 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.