Let be $Y$ a simply connected space. Show that $Y$ doesn't admit covering maps that aren't homeomorphisms, ie, every cover space of $Y$ is trivial ($I\times Y$, with $I$ a discrete space).
So, I know that if $f:X\to Y$ is a cover map and $Y$ is a connected space, then the cardinal of $p^{-1}(y)$, for each $y\in Y$, is constant, and that a homeomorphism between $X$ and $Y$ is a cover maps such that $\# p^{-1}(y)=1$, for every $y\in Y.$
Any idea?
Best Answer
If $Y$ is simply connected then it admits universal covering space trivially $(Y,id)$. You should know that if a space $Y$ has universal cover then this is a covering space of all covering spaces of $Y$. Thus, if $(X,p)$ is a covering space of $Y$ then there exits a homomorphism $\psi$ (which is also a cover map) between $X$ and $Y$ such that $p \circ \psi=id$ (imagine the diagram). By the definition of cover map, $\psi$ and $p$ are continuous and surjective; with the above condition you observe that $p$ is also injective. Moreover $p$ is a cover map then it is a local homeomorphism which implies it is an open map.
Recap: we have found that $p$ is an open continuos bijective map, hence it is a homeomorphism.