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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?

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    $\begingroup$ You have wrote in a perfect way $\endgroup$ – Federico Fallucca Dec 27 '19 at 8:18
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For problem 1, your argument is correct but it wouldn't hurt to include a proof of the fact that a continuous bijective open map is a homeomorphism, as this is not entirely trivial.

For problem 2, your argument is correct (it is a rather trivial question to ask, though).

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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.

The second is completely trivial when embedding is defined (as is common) as a homeomorphism onto its image. Your argument is fine.

Of course an embedding $f: X \to Y$ need not be open or closed but $f'$ always is.

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