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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.
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.
