Lemma 5.7.7. Let $f : X \to Y$ be a continuous map of nonempty topological spaces. Assume that (a) $Y$ is connected, (b) $f$ is open and closed, and (c) there is a point $y\in Y$ such that the fiber $f^{-1}(y)$ is a finite set. Then $X$ has at most $|f^{-1}(y)|$ connected components. Hence any connected component $T$ of $X$ is open and closed, and $f(T)$ is a nonempty open and closed subset of $Y$, which is therefore equal to $Y$.

**Proof.**
If the topological space $X$ has at least $N$ connected components for some $N \in \mathbf{N}$, we find by induction a decomposition $X = X_1 \amalg \ldots \amalg X_ N$ of $X$ as a disjoint union of $N$ nonempty open and closed subsets $X_1, \ldots , X_ N$ of $X$. As $f$ is open and closed, each $f(X_ i)$ is a nonempty open and closed subset of $Y$ and is hence equal to $Y$. In particular the intersection $X_ i \cap f^{-1}(y)$ is nonempty for each $1 \leq i \leq N$. Hence $f^{-1}(y)$ has at least $N$ elements.
$\square$

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## Comments (1)

Comment #630 by Wei Xu on

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