# 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 $D^n$, $n=1,3,4,5...$?

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maybe you mean a cover $p:X\to D^2$... – yoyo Dec 29 '11 at 15:43
@DylanWilson It's not clear to me, Dylan, how you can get a covering map from $D^2$ to a manifold without boundary, since $D^2$ has a boundary and for some $z$ on the boundary, there is no neighborhood of $p(z)$ which is homemorphic with an open Euclidean disk. – Thomas Andrews Dec 29 '11 at 16:01
Sorry- I was confused and was thinking of the open unit disk :) – Dylan Wilson Dec 29 '11 at 16:17
Meanwhile, here's a fun proof: If there are multiple points in a fiber there is some map $D^2 \rightarrow D^2$ switching two of these points. By unique lifting, this implies $f$ has no fixed points. But this contradicts Brouwer's fixed point theorem. The same proof works for all $D^n$. – Dylan Wilson Dec 29 '11 at 16:26
Well if there were a fixed point $x \in D^2$, then choose some other point $y \in D^2$ and a path between $x$ and $y$. $f$ takes this to a path from $x$ to some other point in the same fiber as $y$, but the corresponding path down below hasn't changed (the obvious diagram commutes by definition of deck transformation)... but there's a unique lift of a path below to a path in $D^2$ starting at $x$, so the endpoints must be the same. Thus $f$ is the identity. – Dylan Wilson Dec 29 '11 at 16:57

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.

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$X$ is necessarily path connected since it is the image of a path-connected space - all coverings are surjective. – Thomas Andrews Dec 29 '11 at 17:42
Maybe there are reasons why getting enough sleep is important... I'll edit it. – Dylan Wilson Dec 29 '11 at 18:02
@DylanWilson: could you elaborate please why triviality of deck transformation group implies that $X$ is simply connected? – mathreader Aug 18 '13 at 5:12

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 $\pi_1(X).$

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The image of a path-connected space is path-connected, so $X$ is connected since any covering map is surjective. – Thomas Andrews Dec 29 '11 at 17:44

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.

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