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If $A$ and $B$ are topological spaces and $f:A\to B$ a continuous map and $U$ in $B$ a closed set, why is $f^{-1}(U)$ closed in $A$? I know that preimage of an open set needs to be open.

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$f^{-1}(U)^{c}=f^{-1}(U^c)$, which is open as $U^c$ is open(as $U$ is closed). Hence $f^{-1}(U)$ being complement of an open set, is closed.

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