Let $p$ be a covering space and $X, Y$ be path connected. Show there exists a map $q$ such that $q\circ p=1_{X}$ iff $p$ is a homeomorphism.

Question

Let $p\colon X\rightarrow Y$ be a covering map where $X$ and $Y$ are path connected. Show that there exists a map $q\colon Y\rightarrow X$, such that $q\circ p=1_{X}$ if and only if $p$ is a homeomorphism.

Suppose $p$ is a homeomorphism then there exists a continuous inverse map $p^{-1}\colon Y\rightarrow X$. Since $p$ is a covering space it is a surjective, continuous map. Therefore we can let $q=p^{-1}$ and we get $q\circ p=1_{X}$.

Now suppose that there exists a map $q\colon Y\rightarrow X$ such that $q\circ p=1_{X}$. We know that $p$ is a covering map therefore $p$ is a continuous, surjective map. To show that $p$ is a homeomorphism we have to show that $p$ is also injective. Suppose $p(x_{1})=p(x_{2})$ for some $x_{1},x_{2}\in X$. We need to show that $x_{1}=x_{2}$ and I'm not how to show this?

Any advice would be appreciated. Thank you.

If $p(x_1) = p(x_2)$, what follows for $q(p(x_1))$ and $q(p(x_2))$? — Daniel Fischer, Apr 17 '14 at 20:39
Since the question is trivial if $qp=1$, maybe you meant $pq=1_Y$, i.e. $p$ has a section. We can show that $p$ is a homeomorphism then, see http://math.stackexchange.com/questions/256951/about-covering-maps-and-sections — Stefan Hamcke, Apr 17 '14 at 21:32

score 1 · Answer 1 · answered Apr 19 '14 at 13:01

Let's get this off the unanswered queue.

If $p(x_1)=p(x_2)$ then $q(p(x_1))=q(p(x_2))$ which implies that $q\circ p(x_1)=q\circ p(x_2)$ and so, because $q\circ p$ is assumed to be equal to $1_X$ we get $x_1=x_2$ and so $p$ must be injective, as required.

Let $p$ be a covering space and $X, Y$ be path connected. Show there exists a map $q$ such that $q\circ p=1_{X}$ iff $p$ is a homeomorphism.

1 Answers1