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.
Best Answer
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.