A bijective continuous map is a homeomorphism iff it is open, or equivalently, iff it is closed. [closed]

Wikipedia states that "a bijective continuous map is a homeomorphism if and only if it is open, or equivalently, if and only if it is closed.".
How do we prove this fact?
I can prove the obvious direction, but im unsure how to proceed the other ways

Solution 1:

A bijection $f: X \to Y$ that is closed is also open (and vice versa).

The reason is that if $f$ is a closed bijection and $V$ is an open set in $X$, then the complement $V^c$ of $V$ is closed in $X$, and therefore the image $f(V^c)$ is closed in $Y$ since $f$ is closed. But bijectivity means that $f(V^c) = (f(V))^c$. This means that $f(V)$ is open in $Y$, being the complement of a closed set. Thus, $f$ is an open map as well.

The proof that an open bijection is closed is analogous.

As for why this implies that the map is a homeomorphism: $f$, as given, is continuous by assumption. Further, $f$ has an inverse $f^{-1}$ by bijectivity. Why would this be continuous? Well, we need to know whether $(f^{-1})^{-1}(U)$ is open when $U$ is. But this is the same as $f(U)$, which is open since $f$ is an open map.

Solution 2:

If it's open, then the inverse image by its inverse of any open set is open. This gives the continuity of the inverse and hence the required result.