Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer
$X$ 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 $\pi_1(X).$

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.