Lemma 5.7.5. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that

1. all fibres of $f$ are connected, and

2. a set $T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed.

Then $f$ induces a bijection between the sets of connected components of $X$ and $Y$.

Proof. Let $T \subset Y$ be a connected component. Note that $T$ is closed, see Lemma 5.7.3. The lemma follows if we show that $f^{-1}(T)$ is connected because any connected subset of $X$ maps into a connected component of $Y$ by Lemma 5.7.2. Suppose that $f^{-1}(T) = Z_1 \amalg Z_2$ with $Z_1$, $Z_2$ closed. For any $t \in T$ we see that $f^{-1}(\{ t\} ) = Z_1 \cap f^{-1}(\{ t\} ) \amalg Z_2 \cap f^{-1}(\{ t\} )$. By (1) we see $f^{-1}(\{ t\} )$ is connected we conclude that either $f^{-1}(\{ t\} ) \subset Z_1$ or $f^{-1}(\{ t\} ) \subset Z_2$. In other words $T = T_1 \amalg T_2$ with $f^{-1}(T_ i) = Z_ i$. By (2) we conclude that $T_ i$ is closed in $Y$. Hence either $T_1 = \emptyset$ or $T_2 = \emptyset$ as desired. $\square$

Comment #629 by Wei Xu on

Typos: In the proof, the symbols $p$ should be $f$, or in the statement"$f:X\to Y$" should be "$p:X\to Y$".

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