Fundamental group and doubt about covering map Let $X=\{(x,y)\in \mathbb{R}^2\,\,|\,\, |y|\leq e^x\}$ a subspace of $\mathbb{R}^2$ with the usual topology.
I have to find the fundamental group of $Y=X\setminus\{(0,0)\}$ and determine whether there can be covering maps $p:Y\to S^1$ or $p:Y\to \mathbb{R}^2$.
I am new to this type of exercises, so I can't show anything I've done concretely.
I only know the fundamental groups of $\mathbb{R}^n$ and $S^n$, but I can't conclude anything from this.
Regarding the part related to covering maps, I only studied the theory and this is the first exercise.
Does anyone have an idea about how to procede?
 A: Have you seen that $S^1$ is a deformation retract of $\mathbb{R}^2\setminus\{0\}$? Once you understand this, it is not difficult to cook up a deformation retraction of your space $Y$ onto a circle around the origin homeomorphic to $S^1$ (slightly annoyingly, you cannot directly deformation retract $Y$ to $S^1$, as $S^1$ is not contained in $Y$). Now, you should know that deformation retractions are homotopy equivalences, thus they induce isomorphisms on the fundamental group, and so do homeomorphisms, meaning that the fundamental group of $Y$ is isomorphic to the fundamental group of $S^1$.
As for the question about covering maps, two different observations show that neither any map $Y\to S^1$ nor any map $Y\to \mathbb{R}^2$ can be covering maps. First, no map $Y\to S^1$ can be a covering map, because you would have open subsets of $Y$ mapping homeomorphically to open subsets of $S^1$, which intuitively is not possible as the latter are one dimensional and the former two dimensional: one way to make this precise is to observe that if you delete a point from a connected open subset of $Y$ it stays connected while the same is not true for connected open subsets of $S^1$ (you might want to spend a bit of time working out the details more explicitly). As for maps $Y\to \mathbb{R}^2$, it is a basic fact in covering space theory that the map of fundamental groups induced by a covering map is injective, which cannot be the case as $\pi_1(Y)$ is not trivial, while $\pi_1(\mathbb{R}^2)$ is.
