## profinite spaces are the pro-completion of finite sets

The title sounds tautological, right? Perhaps I'm missing something completely trivial here ...

Assume $X$ is a compact totally disconnected hausdorff space. It is known that $X$ can be written as directed inverse limit of finite discrete spaces $X_i$ with surjective transition maps (i.e. $X$ is profinite). How do you prove that every map from $X$ to a finite discrete space factors through some projection $X \to X_i$?

I know that the fibers of the projections are a basis of the topology of $X$ (not only a subbasis). The corresponding result for profinite groups is true, but I cannot adopt the proof.

Of course you could use Stone duality to reduce the assertions to a completely trivial one (a finite boolean ring in a directed limit of boolean subrings lies in some of these boolean subrings), but I want a direct topological proof.

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Let $f:X \to Z$ be a map to a finite discrete space. Note that each fiber, $f^{-1}(z)$, is both open and closed in $X$. Let $p_i: X \to X_i$ be the projection maps.

Fix some $z \in Z$. Since $f^{-1}(z)$ is open, and the fibers of the maps are a basis, there is an open cover of $f^{-1}(z)$ by sets of the form $p_i^{-1}(x)$, for $x$ in various $X_i$. Since $f^{-1}(z)$ is closed in a compact space, it is compact. So we can take a finite subcover of this cover. Thus, there is some single index $i$ for which $f^{-1}(z)$ is covered by sets of the form $p_i^{-1}(x)$, $x \in X_i$.

Since $Z$ is finite, there is a single $i$ such that, for every $z \in Z$, the fiber $f^{-1}(z)$ is covered by sets of the form $p_i^{-1}(x)$, $x \in X_i$. The map $f$ factors through $X_i$.

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 What is $f^{-1}(x)$ for $x \in X_i$? – Martin Brandenburg Jun 8 2010 at 13:55 Edited, thanks. – David Speyer Jun 8 2010 at 14:14 Ok but I still don't understand the proof at all. a) Why can be choose a single index $i$ in the second paragraph? b) The same in the third paragraph. Yes I have tried to use directedness etc., but it does not work. c) Why does $f$ factor through $X_i$? I mean how do you actually define the map $X_i \to Z$? – Martin Brandenburg Jun 8 2010 at 21:13 Am I missing something? For (a), there are only finitely many indices $j_1$, $j_2$, ..., $j_r$. So, by directedness, there is some $X_i$ which maps to every $X_{j_a}$. Let $x \in X_{j_a}$ and let $q$ be the map $X_i \to X_{j_a}$. So $p_{j_a}^{-1}(x) = \bigcup_{y \in q^{-1}(x)} p_{i}^{-1}(y)$. This shows that a set which is a union of sets of the form $p_{j_a}^{-1}(x)$, for various $x$'s in various $X_{j_a}$, is also a union of sets of the form $p_i^{-1}(x)$ for $x \in X_i$. Similarly for (b). – David Speyer Jun 8 2010 at 21:40 For (c): Suppose that two points, $x_1$ and $x_2$, of $X$ map to the same point of $y \in X_i$. Let $f^{-1}(x_j)=z_j$; we must show that $z_1=z_2$. If not, then $f^{-1}(z_1)$ and $f^{-1}(z_2)$ are disjoint. They are each covered by sets of the form $p_i^{-1}(y')$, for various $y'$ in $X_i$. The only $y'$ for which $p_i^{-1}(y')$ contains $x_1$ is $y$. So $p_i^{-1}(y) \subseteq f^{-1}(z_1)$. But, similarly, $p_i^{-1}(y) \subseteq f^{-1}(z_2)$. This contradicts that $f^{-1}(z_1)$ and $f^{-1}(z_2)$ are disjoint. – David Speyer Jun 8 2010 at 21:49