Lemma 5.7.5. Let f : X \to Y be a continuous map of topological spaces. Assume that
all fibres of f are connected, and
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
Comments (1)
Comment #629 by Wei Xu on
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