Covering map on the unit disk Let $f: D^2 \rightarrow X$ be a covering map. I am trying to show that $f$ must in fact be a homeomorphism. To do so, I believe it suffices to show that $f$ is injective. Moreover, if only one point of $X$ has a finite pre-image, we can use a connectedness argument to show that $f$ is injective on all of $D^2$. So far, I have attempted to prove this using compactness to obtain finitely many open sets of $X$ which cover $X$ and are each evenly covered under $f$, but have been unsuccessful. Any suggestions?
Also, I am wondering how this generalizes to other compact, simply connected spaces. Is the same true if we replace $D^2$ by $D^n$, $n=1,3,4,5...$?
 A: Sorry, I may as well just post this as an answer:
The theorem holds for any $D^n$ (the closed disk).
Proof: It suffices to show that $X$ is simply connected, i.e. that $\pi_1(X) = 0$. By covering space theory, since $D^n$ is simply connected, this is the same as showing that the group of deck transformations of $p: D^2\rightarrow X$ is trivial. But any deck transformation is, among other things, a map $f: D^n \rightarrow D^n$. By Brouwer's theorem this has a fixed point, but this implies that $f$ is the identity.
Indeed, choose $x\in D^2$ fixed by $f$. Choose a path $\Gamma: [0,1] \rightarrow D^n$ from $x$ to any other point $y$. Then $f \circ \Gamma$ is a path in $D^n$ and since $p\circ f = p$ (by the definition of "deck transformation"), $p \circ f \circ \Gamma = p \circ \Gamma$. It follows from this and the fact that $f(x) = x$ that $f \circ \Gamma$ and $\Gamma$ are lifts of the same path with the same starting point, whence they coincide by the unique lifting property, so in particular they have the same endpoint. Thus $f(y) = y$, and the proof is done since $y$ is arbitrary.
A: I think you can do the following assuming $X$ is connected.
a) Use the universal lifting lemma to show that $\pi_1(X)$ is trivial. 
b) Use the fact that the size of the fiber is the index of $f_{\ast}\pi_{1}(D^2)$ in $\pi_1(X).$
A: Also, one can reason like that:
Let $p:D^n\to X$ be a covering. Since $\pi_1(D^2)=1$, it is normal viewed as a subgroup of $\pi_1(X)$, and therefore the covering $p$ is regular, hence the deck transformation group acts transitively on any set $p^{-1}(x)$, $x\in X$. If the covering is $m$-sheeted with $m>1$, then there is a nontrivial element in the deck transformation group, acting on $m$ points of $p^{-1}(x)$. But by the fixed point theorem, any such transformation should have a fixed point in $D^n$, and hence be trivial, as all deck transformations act freely. Hence $m=1$, and $p$ is a homeomorphism.
