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Exercise: Let $f:(X,tau)to (Y,tau_1)$ be a continuous bijection. If $(X,tau)$ is compact and $(Y,tau_1)$ is Hausdorff, prove that $f$ is a homeomorphism.

I tried to prove this on the following way:

First I proved the following Lemma:

Lemma: If $(X,tau)$ and $(Y,tau_1) $are compact Hausdorff spaces and $f:(X,tau)to(Y,tau_1)$ is a continuous mapping then $f$ is a closed mapping.

Proof: If $Ain X$ is compact than it is closed in $(X,tau)$. Then if ${a_n:ninmathbb{N}}$ is an arbitrary sequence in A then by the compactness there is a subsequence that converges in A such that $lim_{ntoinfty}a_{in}=a$ where $a in A$. By continuity of $f$, $lim_{ntoinfty} f(a_{in})=f(a)$ so that $f(a)in f(A)$. So $f(A)$ is compact since the space $(Y,tau_1)$ is compact then $f(A)$ is closed. So $f$ is a closed mapping.

In the Exercise the function is continuous so if $Bintau_1$ then $f^{-1}(B)intau$, now it is left to show that $f$ send open sets to open sets. This is where my problem begins:

Compactness is going to be preserved by continuity of $f$, then $(Y,tau_1)$ must be compact as every image of a subset of $(X,tau)$ that would imply that $f$ is a closed mapping by the Lemma. If $C$ is a closed set in $X,tau$ then $f(Xsetminus C)=Xsetminus f(C)$ which must be open. However I am not certain about this last step.

Question:

How should I solve the question? Is my proof right?

Thanks in advance!

The continuous image of a compact space is compact. We don't need sequences to see this; in fact sequences don't even suffice to see it, in general. The definition of compactness is by open covers, so use that:

If $f:X to Y$ is continuous, $A subseteq X$ is compact, then consider an open cover $O_i, i in I$ of $f[A]$. Then $f^{-1}[O_i], i in I$ is a cover of $A$ (by basic set theory) and an open cover as $f$ is continuous. So finitely many $f^{-1}[O_i], i in F$ (so $F subseteq I$ finite) exist that also cover $A$ and again simple set theory tells us that the $O_i, i in F$ is a finite subcover of the original cover for $f[A]$. Hence $f[A]$ is compact.

The lemma then follows from the basic fact that if $Y$ is Hausdorff, and $B subseteq Y$ is compact, then $B$ is closed in $Y$. This too is shown using open covers and the definition of Hausdorffness. Plenty of proofs can be found online.

Now if a bijection $f: X to Y$ is closed, this is the same as saying its inverse map $g: Y to X$ is continuous: $g$ is continuous iff $g^{-1}[C]$ is closed for all closed $C subseteq X$. And $g^{-1}[C] = f[C]$ because $g$ is the inverse of the bijection $f$. As $f$ is a closed map by the lemma, you're done.

Your proof that $f$ is closed is (a little) bad. Because you suppose to begin with a closed $A$ not a compact $A$. However this is not a big deal because a closed subset of a compact set is compact. Moreover the fact that $f(A)$ is compact in a compact space $Y$ doesn't necessarily means that is closed$^1$. For this you need the fact that $Y$ is Hausdorff (compact set in an Hausdorff space is closed).

About the second, you're right. A bijection that is also closed is necessarily open because $f(Xbackslash C) = Ybackslash f(C)$.

To prove this you can show that each of the sets is included in the other. Let $yin f(Xbackslash C)$ then clearly $yin Y$ but $ynot in f(C)$ because $f$ is injective.

On the other hand if $yin Y$ then since $f$ is onto there exists $xin X$ such that $f(x)=y$. Moreover if $ynotin f(C)$ then $xnotin C$ again because $f$ is injective.

1. For example if $Y$ is equipped with the trivial topology it is always compact (and every subset of it is compact) but no non-trivial subset is closed.