4
$\begingroup$

Write $\mathbf{Stone}$ for the category of compact, Hausdorff totally disconnected spaces, and write $U:\mathbf{Stone}\to\mathbf{Set}$ for the forgetful functor.

I think it must be likely that $U$ does not preserve colimits, but I can not see why - I am confident that it preserves coproducts, but I don't know about coequalizers.

$\endgroup$

1 Answer 1

4
$\begingroup$

The functor $U$ preserves neither infinite coproducts nor coequalizers. For instance, if you take countably infinitely many $1$-point spaces, then their coproduct in $\mathbf{Stone}$ is $\beta\mathbb{N}$, which has uncountably many points.

For a coequalizer that is not preserved, let $K\subset[0,1]$ be the Cantor set and denote the intervals that are removed from $[0,1]$ to form $K$ by $(a_n,b_n)$. Consider two maps $f,g:\beta\mathbb{N}\to K$, where $f(n)=a_n$ for each $n\in\mathbb{N}$ and $g(n)=b_n$ for each $n\in\mathbb{N}$. Also consider the map $h:K\to [0,1]$ defined by $$h\left(\sum_{n=1}^\infty c_n3^{-n}\right)=\sum_{n=1}^\infty \frac{c_n}{2}2^{-n},$$ where each $c_n$ is either $0$ or $2$. Then $h$ is continuous and surjective (and hence a quotient map since the spaces involved are compact Hausdorff). Moreover, $h$ is injective except that $h(a_n)=h(b_n)$ for all $n$. Since $\mathbb{N}$ is dense in $\beta\mathbb{N}$, it follows that $h\circ f=h\circ g$, and thus that the equivalence relation on $K$ generated by $f$ and $g$ is just that $a_n\sim b_n$ for each $n$. In particular, $h$ is the quotient map by this equivalence relation.

It follows that as a map of sets, $h:K\to [0,1]$ is the coequalizer of $Uf$ and $Ug$.

On the other hand, the coequalizer of $f$ and $g$ in $\mathbf{Stone}$ is the Stone-ification of the coequalizer of $f$ and $g$ in $\mathbf{Top}$, which is $h:K\to [0,1]$ (since $h$ is the quotient map by the equivalence relation generated by $f$ and $g$). Since $[0,1]$ is connected, its Stone-ification has just one point. So the coequalizer of $f$ and $g$ in $\mathbf{Stone}$ is a $1$-point space.

$\endgroup$
3
  • $\begingroup$ Thanks! Am I right in thinking that the infinite coproduct counterexample does not work anymore if we instead ask the same question for the forgetful functor $U_*$ from pointed Stone spaces to pointed sets? $\endgroup$ Commented Sep 12, 2016 at 5:16
  • $\begingroup$ Well, that particular counterexample doesn't work, but you can get another one by just taking a coproduct of infinitely many $2$-point spaces (the coproduct in pointed Stone spaces will just be $\beta\mathbb{N}$ with a disjoint basepoint added). $\endgroup$ Commented Sep 12, 2016 at 5:17
  • $\begingroup$ chuckle I realize that it’s the specific datum that you need, but saying that $\beta\Bbb N$ has uncountably many points does seem a bit like saying that water is wet! $\endgroup$ Commented Sep 12, 2016 at 6:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .