# Topological embeddings

Problem 1: A continuous injective map that is either open or closed is a topological embedding.

Solution:Without loss of generality suppose $$f:X\rightarrow Y$$ is a continuous injective open map. Then $$f: X\rightarrow f(X)$$ is a continuous bijection. To show that it is a homeomorphism, it suffices to show that $$f$$ onto its image is open. Let $$U$$ be open in $$X$$ so by assumption, $$f(U)$$ is an open subset of $$Y$$. Since $$f(U)\subseteq f(X)$$, $$f(U)= f(U)\cap f(X)$$, which is open in $$f(X)$$. Hence $$f$$ onto its image is a homeomorphism. Thus $$f$$ is a topological embedding.

Problem 2: A surjective topological embedding is a homeomorphism

Solution:Suppose $$f:X\rightarrow Y$$ is a surjective topological embedding, so $$f:X\rightarrow f(X)$$ is a homeomorphism, but $$f(X)=Y$$ since $$f$$ is surjective, so $$f:X\rightarrow Y$$ is a homeomorphism.

Are the solutions correct?

• You have wrote in a perfect way – Federico Fallucca Dec 27 '19 at 8:18

You did not handle the closed case for problem 1, but the idea is fine. If $$f$$ is open, $$f': X \to f[X]$$ (so the codomain restricted, $$f'(x)=f(x)$$ for all $$x$$) is a 1-1 open continuous bijection and so a homeomorphism.
The same goes for closed: a 1-1 closed continuous bijection is a homeomorphism too. And $$f$$ closed trivially implies $$f'$$ closed too.
Of course an embedding $$f: X \to Y$$ need not be open or closed but $$f'$$ always is.