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

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.

• 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. Jun 27 '19 at 15:39

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