The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.
Prove that if $X$ and $Y$ are compact Hausdroff space and $f:X\rightarrow Y$ is a continuous surjection, then $f$ is a quotient map.
This is what I understand:
- A space $X$ is compact provided that every open cover of $X$ has a finite subcover.
- A Hausdorff Space is a space in which distinct points have disjoint neighborhoods.
- Since $X$ and $Y$ are two topological spaces and $f: X\rightarrow Y$ a surjective map. The map $f$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. Equivalently, subset $U$ of $Y$ closed in $Y$ if and only if $f^{-1}(U)$ closed in $X$.
From what we are given from the problem statement:
- The map $f:X\rightarrow Y$ is surjective. Thus for any $B\subset Y$, we have $f^{-1}(Y-B)=X-f^{-1}(B)$.
- The map $f: X\rightarrow Y$ is continuous. In other words, $U$ open in $Y\Rightarrow$ $f^{-1}(U)$ open in $X$.
- All I need to show is $U\subseteq Y$, $f^{-1}(U)$ open in $X\Rightarrow U$ open in $Y$.
My question is, how would I use the fact that $X$ and $Y$ are compact Hausdroff spaces to prove point 3? Now every compact space is locally compact, meaning for every $x\in X$, there is an open set containing $x$ for which $\overline{U}$ is compact.
I'm probably incorrect with many of the following implications but it's a really rough start with what I had in mind.
Let $A$ be a neighborhood of $a\in X$ that is disjoint from other neighborhoods of distinct points from $a$ in $X$. Now, $f(A)$ is a neighborhood of some $b$ in $Y$. Then $\overline{f({A})}$ is a compact subset of $Y$.
Granted that last part is probably a big jump that I can't assume. My idea was to some how eventually show that $f(A)\subset Y$, $f^{-1}(f(A))$ open in $X\Rightarrow f(A)$ open in $Y$. It's just a rough idea. If I am on to something with what I just stated, how would I go about constructing the proof? Any suggestions?
Sorry for the long read. If I need to clarify on some ideas and thoughts I presented, let me know. I sincerely thank you for taking the time to read this question. I greatly appreciate any assistance you may provide.
Best Answer
Obviously, by continuity of $f$, if $V$ is closed in $Y$ then $f^{-1}(V)$ is closed in $X$, so we only have to check the other direction.
Since $f$ is surjective we have that $f(f^{-1}(V))=V$ for every set $V$. Assume now that $W=f^{-1}(V)$ is closed in $X$. Then since $X$ is compact we get $W$ compact, and since $f$ is continuous $f(W)=V$ is compact and therefore, since $Y$ is Hausdorff, also closed.