# Homeomorphism of Topological Spaces

I'm attempting to solve a problem that is as follows,

"If $f:X\rightarrow Y$ is a covering map and $X$ is path connected and $Y$ is simply connected then prove that $f$ is a homeomorphism of Topological spaces $X$ and $Y$"

My solution is as follows:

Let $y_0\in Y$. Since $f$ is a covering map, $y_0$ is contained in an open $U_{y_0}$ which is evenly covered. Thus, $f^{-1}(U_{y_0})=\cup_{a\in A}V_a$ where the $V_a$'s are open in $X$ and disjoint and each one is mapped homeomorphically to $U_{y_0}$. Now, assume that there are $x_0,x_1$ with $x_i\in V_{a_i}, i=0,1$ and $f(x_i)=y_0$. Since, $X$ is path connected we have a continuous $x:[0,1]\rightarrow X$ with $x(0)=x_0, x(1)=x_1$ so we take $y=f\circ x$ then this defines a loop in $Y$. Since $Y$ is simply connected, $f\circ x$ is homotopic to a point.

I'm not sure where to go from here or even if the last few lines are going in the right direction so any hints would be appreciated. Thanks.

• Do you know the path lifting lemma and the homotopy lifting lemma? – Lee Mosher Nov 15 '16 at 18:18
• I don't think so, could you give the formal statement? – Crunch Nov 15 '16 at 18:26
• It is something you can look up in most topology books, I recommend Munkres' book. – Lee Mosher Nov 15 '16 at 20:00

## 1 Answer

This is perfectly correct. You can finish by the following : lift the homotopy of $f \circ x$ to the covering space and prove that $x_1 = x_0$, which gives a contradiction.