# Homomorphism from colimit in $\mathsf{BA}$ to other Boolean algebra injective?

Setting. Let $\mathcal{X}$ be a Stone space with underlying space $X$ and $\mathbb{T}$ an endofunctor on $\mathsf{Set}$, the category of sets and functions. Suppose $\mathbb{T}$ sends finite sets to finite sets (i.e., it restricts to $\mathsf{FinSet}$). Let $\{ q_i : \mathcal{X} \to \mathcal{Y}_i \mid i \in I \}$ be the collection of finite quotients of $\mathcal{X}$. A topology on a finite set is a Stone topology iff it is discrete, so we can view $\mathbb{T}$ as an endofunctor on $\mathsf{FinStone}$. Let $\mathcal{Z}$ be the limit of the full diagram (*) $\mathbb{T}\mathcal{Y}_i$ (where $i$ ranges over $I$) in $\mathsf{Stone}$, the category of Stone spaces and continuous maps.

Viewed as sets, we have for each $\mathcal{Y}_i$ a map $\mathbb{T}q_i : \mathbb{T}X \to \mathbb{T}\mathcal{Y}_i$. Now look at it dually, in the category $\mathsf{BA}$ of Boolean algebra and homomorphisms. Then $\operatorname{Clop}\mathcal{Z}$ is the colimit over the diagram dual to (*). For each $i$ we have a Boolean algebra homomorphism $(\mathbb{T}q_i)^{-1} : \operatorname{Clop}(\mathbb{T}\mathcal{Y}_i) \to \mathbb{PT}X$, where the latter is the powerset Boolean algebra. Clearly, if $g : \operatorname{Clop}(\mathbb{T}\mathcal{Y}_i) \to \operatorname{Clop}(\mathbb{T}\mathcal{Y}_j)$ is a homomorphism, we have $(\mathbb{T}q_i)^{-1} = (\mathbb{T}q_j)^{-1} \circ g$. So by the definition of a colimit, there exists a unique homomorphism $f : \operatorname{Clop}\mathcal{Z} \to \mathbb{PT}X$.

Aim. I want to prove that the Boolean algebra $\operatorname{Clop}(\mathcal{Z})$ is isomorphic to the sub-Boolean algebra of $\mathbb{PT}X$ generated by $(\mathbb{T}q_i)^{-1}(a)$, where $i \in I$ and $a \subseteq \operatorname{Clop}(\mathbb{T}\mathcal{Y}_i)$.

I think the following suffices:

Question. Is $f$ injective?

• To be clear, when you say "full diagram", you mean the diagram obtained by applying $\mathbb{T}$ to the diagram of maps between the sets $\mathcal{Y}_i$ which are compatible with the maps $q_i$? And you require $g$ to be a map in this diagram? Otherwise I don't understand the claim which you say is "Clearly" true. – Eric Wofsey Apr 12 '18 at 4:34

The key observation here is that each $q_i$ has a section. So let's fix maps $r_i:\mathcal{Y}_i\to\mathcal{X}$ such that $q_ir_i=1_{\mathcal{Y}_i}$ for each $i$. So now suppose we have some element $z\in\operatorname{Clop}\mathcal{Z}$ such that $f(z)=0$. Then $z$ comes from some element $y\in\operatorname{Clop}(\mathbb{T}\mathcal{Y}_i)$ for some $i$ (since our diagram with colimit $\operatorname{Clop}\mathcal{Z}$ is filtered), and $f(z)=(\mathbb{T}q_i)^{-1}(y)$. But $(\mathbb{T}q_i)^{-1}$ is injective, since it has left inverse $(\mathbb{T}r_i)^{-1}$. Thus $(\mathbb{T}q_i)^{-1}(y)=0$ implies $y=0$ and so $z=0$, and $f$ is injective.
(Your "aim" does indeed then follow, since the subalgebra you describe is exactly the image of $f$.)
• This may be a silly question, but how does the fact that our colimit is filtered prove that there exists a $z$ such that $f(z) = (\mathbb{T}q_i)^{-1}(y)$? – Math Student 020 Jun 12 '18 at 15:33
• (Then, we have $f(z)=(\mathbb{T}q_i)^{-1}(y)$ by definition of $f$, since $f$ is the unique map whose composition with each inclusion into the colimit is $(\mathbb{T}q_i)^{-1}$.) – Eric Wofsey Jun 12 '18 at 15:50