Let p:$\displaystyle \overset{\sim}{X}$ --> X be a covering and let f, g: Y --> $\displaystyle \overset{\sim}{X}$ 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 p:$\displaystyle \overset{\sim}{X}$ --> X be a covering and let f, g: Y --> $\displaystyle \overset{\sim}{X}$ 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.
I am just a little confused. Usually the notation $\displaystyle X/\sim$ is used to denote a quotient space (by some equivalence relation $\displaystyle \sim$) but I think you might just mean $\displaystyle \overset{\sim}{X}$.
Ok, so I have a visual here is what we have
Right?
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 $\displaystyle f\oplus g:Y\to\overset{\sim}{X}\times\overset{\sim}{X}:x\m apsto(f(x),g(x))$.
Hmm, obviously the tricks going to have to do something with if $\displaystyle x\in A(f,g)=\left\{y\in Y:f(y)=g(y)\right\}$ then the fibers of $\displaystyle p(f(x)),p(g(x))$ are going to be same.
I'm sorry for getting your hopes, I'll try to figure it out.
Let $\displaystyle K=\{y \in Y| f(y) = g(y)\}$ and let $\displaystyle y_0$ be an arbitrary point of the closure of $\displaystyle K$. Let $\displaystyle x_0 = pf(y_0)=pg(y_0)$. We shall show that $\displaystyle y_0 \in K$, which implies that $\displaystyle K$ is a closed set of Y.
Suppose to the contrary that $\displaystyle y_0 \notin K$. Then $\displaystyle f(y_0) \neq g(y_0)$. Let U be an open neighborhood of $\displaystyle x_0$. Then $\displaystyle p^{-1}(U)$ contains $\displaystyle f(y_0)$ and $\displaystyle g(y_0)$, respectively. Let V and W be open components of $\displaystyle p^{-1}(U)$ such that $\displaystyle f(y_0) \in V$ and $\displaystyle g(y_0) \in W$, where V and W are disjoint open sets. By hypothesis, f and g are continuous, so there exist open neighborhoods $\displaystyle M$ of $\displaystyle y_0$ such that $\displaystyle f(M) \subset V$ and $\displaystyle g(M) \subset W$. This contradicts that $\displaystyle y_0$ is a point of the closure of $\displaystyle K$. Thus $\displaystyle y_0 \in K$ and K is a closed set of Y.
To show that K is open in Y, you need to show that for any $\displaystyle y_0 \in K$, there exists an open set contains $\displaystyle y_0$ and is contained in K. Let V and W be an open component of $\displaystyle p^{-1}(U)$ such that $\displaystyle f(y_0) \in V$ and $\displaystyle g(y_0) \in W$, where U is an open neighborhood of $\displaystyle x_0=pf(y_0)=pg(y_0)$. Since $\displaystyle y_0 \in K$, we have $\displaystyle V=W$. Thus K is an open set of Y.