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This is a problem from Introduction to Topology: Pure and Applied by Colin Adams and Robert Franzosa.

Problem

"Prove that a bijection $f:X→Y$ is a homeomorphism if and only if $f$ and $f^{-1}$ map closed sets to closed sets."

Definition

"We can paraphrase the definition of homeomorphism by saying that $f$ is a homeomorphism if it is a bijection on points and a bijection on the collections of open sets making up the topologies involved. Every point in $X$ is matched to a unique point in $Y$, with no points in $Y$ left over. At the same time, every open set in $X$ is matched to a unique open set in $Y$, with no open sets in $Y$ left over."

Thoughts

Let $f:X→Y$ be a bijection.

Suppose $f^{-1}$ does not map the closed $C'$ to a closed set C. Then $f^{-1}$ does not map the open set $Y-C'$ to an open set $X-C$. Then $f:X→Y$ is not a homeomorphism.

Suppose $f$ maps all closed sets $C$ to all closed sets $C'$, and $f^{-1}$ maps all closed sets $C'$ to all closed sets $C$. Then $f$ maps all open sets $X-C$ to open sets $Y-C'$, and $f^{-1}$ maps all open sets $Y-C'$ to open sets $X-C$. Then $f:X→Y$ is a homeomorphism.

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    $\begingroup$ Is that the real definition of homeomorphism, that you got? Normally a function $f:X\to Y$ is called a homeomorphism if $f$ is a continuous bijection and $f^{-1}$ is continuous. The property that $f$ and $f^{-1}$ map closed sets to closed sets is equivalent to $f$ and $f^{-1}$ beeing continuous. $\endgroup$
    – Cornman
    Jun 27 '19 at 15:39
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The maps $f$ and $f^{−1}$ are closed iff they are continuous:

Suppose $f$ is a homeomorphism and let $A \subset X$ be a closed set. We get that $f(A) = (f^{-1})^{-1}(A) \subset Y$ is closed since $f^{-1}$ is continuous. Analogously $f^{-1}$ is closed.

Suppose $f$ and $f^{-1}$ are closed, and let $B \subset Y$ be a closed set. Now we have that $f^{-1}(B) \subset X$ is closed as $f^{-1}$ is a closed map. Therefore $f$ is continuous. Analogously $f^{-1}$ is continuous.

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Hint: $f$ is closed iff $f^{-1}$ is continuous.

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