# Thread: [SOLVED] Algebraic Topology: Covering Spaces

1. ## [SOLVED] Algebraic Topology: Covering Spaces

Let p: $\overset{\sim}{X}$ --> X be a covering and let f, g: Y --> $\overset{\sim}{X}$ be two continuous maps with pf=pg. Prove that thee set of points in Y for which f and g agree is an open and closed subset of Y.

2. Originally Posted by JJMC89
Let p:X~ --> X be a covering and let f, g: Y --> X~ be two continuous maps with pf=pg. Prove that thee set of points in Y for which f and g agree is an open and closed subset of Y.
I am just a little confused. Usually the notation $X/\sim$ is used to denote a quotient space (by some equivalence relation $\sim$) but I think you might just mean $\overset{\sim}{X}$.

Ok, so I have a visual here is what we have

$\begin{matrix}&space;Y&space;&\xrightarrow[\quad\quad\quad]{f}&space;&\overset{\sim}{X}&space;\\&space;^{g}\bigg\downarrow&&space;&\bigg\downarrow{^{p}}&space;\\&space;\overset{\sim}{X}&\xrightarrow[\quad\quad\quad]{p}&space;&X&space;\end{matrix}$

Right?

3. Correct.
I edited my post to show that it is $\overset{\sim}{X}$.

4. Originally Posted by JJMC89
Correct.
I edited my post to show that it is $\overset{\sim}{X}$.
Hmm...I thought I had it but I don't. Let me try to think of something else. A common trick when doing things like this is to consider that map $f\oplus g:Y\to\overset{\sim}{X}\times\overset{\sim}{X}:x\m apsto(f(x),g(x))$.

Hmm, obviously the tricks going to have to do something with if $x\in A(f,g)=\left\{y\in Y:f(y)=g(y)\right\}$ then the fibers of $p(f(x)),p(g(x))$ are going to be same.

I'm sorry for getting your hopes, I'll try to figure it out.

5. Originally Posted by Drexel28
I'm sorry for getting your hopes, I'll try to figure it out.
Thanks

6. Originally Posted by JJMC89
Let p: $\overset{\sim}{X}$ --> X be a covering and let f, g: Y --> $\overset{\sim}{X}$ be two continuous maps with pf=pg. Prove that thee set of points in Y for which f and g agree is an open and closed subset of Y.
Let $K=\{y \in Y| f(y) = g(y)\}$ and let $y_0$ be an arbitrary point of the closure of $K$. Let $x_0 = pf(y_0)=pg(y_0)$. We shall show that $y_0 \in K$, which implies that $K$ is a closed set of Y.
Suppose to the contrary that $y_0 \notin K$. Then $f(y_0) \neq g(y_0)$. Let U be an open neighborhood of $x_0$. Then $p^{-1}(U)$ contains $f(y_0)$ and $g(y_0)$, respectively. Let V and W be open components of $p^{-1}(U)$ such that $f(y_0) \in V$ and $g(y_0) \in W$, where V and W are disjoint open sets. By hypothesis, f and g are continuous, so there exist open neighborhoods $M$ of $y_0$ such that $f(M) \subset V$ and $g(M) \subset W$. This contradicts that $y_0$ is a point of the closure of $K$. Thus $y_0 \in K$ and K is a closed set of Y.

To show that K is open in Y, you need to show that for any $y_0 \in K$, there exists an open set contains $y_0$ and is contained in K. Let V and W be an open component of $p^{-1}(U)$ such that $f(y_0) \in V$ and $g(y_0) \in W$, where U is an open neighborhood of $x_0=pf(y_0)=pg(y_0)$. Since $y_0 \in K$, we have $V=W$. Thus K is an open set of Y.