Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was thinking about my previous question and thought about going the other way around. Assume we are given a space $Y$ and $Y$ covers $X$, then how much can be said about $X$? The most trivial property would be that if $Y$ is compact, then $X$ would also be compact. In other words, I would like to classify all spaces that a particular space can cover.

Lets pick for example the unit disk $D^1$. Assuming $D^1\to X$ is a cover and $X$ is a CW-complex, then the Euler characteristics gives $\chi(D^1)=1$, so $\chi(X)=1$. The classification theorem would then imply that $X\approx D^1$, since the universal cover of a simply-connected space is just itself.

The second easiest example is probably $\mathbb{C}P^2$. Now the Euler characteristic is $2$, so if it covers a CW-complex $X$, then simple-connectivity would force the cover to have degree $1$ or $2$. In the first case, we would just have the space itself. If the cover has degree $2$ I'm not completely sure what to do, since the only thing we seem to know is that $\pi_1(X)=\mathbb{Z}_2?$

If we drop the assumption that $X$ is a CW-complex, then can anything meaningful be said except that higher homotopy groups are equal?

share|improve this question
    
From a prior question, any covering map from $D^n$ is a homeomorphism: math.stackexchange.com/questions/94936/… –  Thomas Andrews Jan 17 '12 at 14:50
    
Based on the proof given there, it would then suffice to show that any homeomorphism $f:\mathbb{C}P^2\to \mathbb{C}P^2$ has a fixed point, since that would show that the fundamental group of $X$ must be trivial (equals group of deck transformations) and the cover would have to have just 1-sheet. Do we always know such a map has a fixed point? –  pki Jan 17 '12 at 14:57
    
I didn't mean to imply that that proof would extend to $\mathbb CP^2$, only that it covered the $D^n$ case. –  Thomas Andrews Jan 17 '12 at 15:01
    
@ThomasAndrews: You don't have to imply it, since it does extend. ;) The only thing we need is to show that the fundamental group is trivial which forces the number of sheets to be $1$. This clearly follows if the group of deck transformations is trivial, which is the method of proof that was used in that question. –  pki Jan 17 '12 at 15:04
    
Finding such a fixed point seems to follow from the Lefshetz fixed-point theorem... It seems like my two examples are then covered. How much can be said about the general case? Any other strategies to deduce information about the base space? –  pki Jan 17 '12 at 15:07
show 2 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.