Welcome to our community

Be a part of something great, join today!

Problem of the Week #261 - Dec 05, 2017

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
Here is this week's POTW:

-----
If $\phi : A \to B$ is a local homeomorphism from a compact space $A$ to a connected Hausdorff space $B$, show that $\phi$ is surjective and the fibers of $\phi$ over the points of $B$ are finite sets.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
This week's problem was solved correctly by Opalg and GJA . You can read GJA 's solution below.


Begin Proof

To prove that $\phi$ is onto, we will show that $\phi(A)$ is both a closed and open (i.e., clopen) subset of the connected set $B$.

Proving $\phi(A)$ is closed

Since $\phi:A\rightarrow B$ is a local homeomorphism, it is necessarily continuous. It then follows that $\phi(A)$ is a compact subset of $B$ since the continuous image of a compact set is compact. As compact subsets of Hausdorff spaces are necessarily closed, we have that $\phi(A)$ is a closed subset of $B$.

Proving $\phi(A)$ is open

Using the local homeomorphism property of $\phi$, there is a collection of open sets $\{V_{x}\}_{x\in A}$ such that $\phi|_{V_{x}}$ is a homeomorphism of $V_{x}$ onto $\phi(V_{x})$. In particular, $\phi(V_{x})$ is an open subset of $B$ for each $x\in A.$ Using the fact that functions distribute over unions and that the union of open sets is open, we have that

$\phi(A)=\phi\left(\cup_{x\in A} V_{x} \right)=\cup_{x\in A} \phi(V_{x})$

is open in $B$.

Since $B$ is connected, it follows from the clopenness of $\phi(A)$ that $\phi(A)=B;$ i.e., $\phi:A\rightarrow B$ is surjective.

Proving the fibers of $\phi$ are finite

Let $p\in B$. Since $B$ is Hausdorff, $\{p\}$ is a closed subset of $B$. Since $\phi$ is continuous, $K=\phi^{-1}(\{p\})$ is closed in $A$. As a closed subset of the compact set $A$, $K$ is compact.

Now, the key to this portion of the argument is to note that the open sets involved in the local homeomorphism property of $\phi$ can only intersect $K$ at a single point. That is, let $x\in K$ and, as above, let $V_{x}$ denote an open set of $A$ containing $x$ coming from the local homeomorphism property of $\phi$. Then $V_{x}\cap\left( K\backslash\{x\}\right)=\emptyset$ for, otherwise, if $q\in V_{x}\cap\left( K\backslash\{x\}\right)$, then $\phi(q)=\phi(x)=p$ by definition of $K$, contradicting the homeomorphism property (i.e., failure to be injective) of $\phi|_{V_{x}}.$ Having noted this, if $K$ were infinite, then $\{V_{x}\}_{x\in K}$ would be an infinite open cover of $K$ that does not admit a finite subcover (since $V_{x}$ intersects $K$ at $x$ only), contradicting the fact that $K$ is a compact subset of $A$. Hence, $K=\phi^{-1}(\{p\})$ is finite. Since $p\in B$ was chosen arbitrarily, the fibers of $\phi$ are finite subsets of $A$.
 
Status
Not open for further replies.