Topology: Continuous bijective function, domain = covering compact

Question

Final in real analysis coming up. I could really use some help.

If a function f from one set M to another set N is a continuous bijection and M is covering compact, can anything in general be said about f being a homeomorphism?

Thanks in advance!

Emma

Answer 1

You many want to assume that $N$ is Hausdorff in order to conclude that $f$ is a homeomorphism. To show that $f$ is homeomorphic, it suffices to show that it maps closed subset to closed subset (i.e. $f^{-1}$ is continuous as well). Any closed subset in $M$ is compact, so it get sent to a compact subset of $N$. If $N$ is Hausdorff, then this compact subset is closed.

Answer 2

As taken from a comment above, $f$ is a function between metric spaces. In metric spaces each compact set is closed which can be seen like this (if the Hausdorff property shall not be used): Each compact set in a metric space is complete and totally bounded. Complete sets are closed.

Now $f^{-1}$ is continuous if and only if preimages of closed sets are closed, that means $(f^{-1})^{-1}(A)$ is closed for each closed $A\subset M$. This is equivalent to $f(A)$ is closed for each $A\subset M$. But if $A\subset M$ is closed, it is compact as closed subset of a compact space. Hence, using continuity of $f$, $f(A)$ is compact, hence $f(A)$ is closed as shown above.

(This is by the way exactly the same answer as the one of Alek Fok, just without using the word Hausdorff and probably a little more detail. You might aswell accept his answer if you are content with this one.)

Answer 3

$f$ is a homeomorphism iff $f$ is a continuous bijection, and $f^{-1}$ is continuous. So, it remains to prove that $f^{-1}$ is continuous.

To explain what was going on in Alex's answer:

A topological space $(X,\tau)$ is said to have the Hausdorff property if: for any $x,y \in X$, $x\neq y$, we can separate $x$ and $y$ by two disjoint open subsets. That is, there are $U,V \subset X$ such that: $U,V \in \tau$, $x \in U$, $y \in V$, and $U \cap V = \varnothing$.

Any metric space $(X,d)$ is a Hausdorff space because: let $x,y \in X$, $x \neq y$. Let $\epsilon = d(x,y) > 0$. Let $U = B_d(x,\epsilon/3)$ and $V = B_d(y, \epsilon/3)$.

So, we don't need to worry about that property as long as we are living in a metric space.

Back to the question: to show $f^{-1}$ is continuous, it suffices to prove that the inverse image by $f^{-1}$ of any closed subset of $M$ is a closed subset of $N$. That is, the image by $f$ of any closed subset is closed.

Let $C$ be closed in $M$. Since $M$ is compact, $C$ becomes compact. We know that the continuous image of a compact set is compact. Then, $f(C)$ is compact in $N$. Now, since $N$ is a metric space and hence Hausdorff, $f(C)$ becomes closed in $N$. This completes the proof.