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If $q: E\rightarrow X$ is a covering map that has a section $(i.e. f: X\rightarrow E, q\circ f=Id_X)$ does that imply that $E$ is a 1-fold cover?

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Does it have only one section? –  Andy Dec 12 '12 at 9:07
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Well, what if $E = X \amalg X$? –  Zhen Lin Dec 12 '12 at 9:31
    
@ZhenLin: I forgot to add that $E$ has to be connected...because that obviously would not hold in case $E$ is not connected, as you pointed out. –  hesiar Dec 12 '12 at 16:31
    
@Andy I'm not sure how that makes a difference? –  hesiar Dec 12 '12 at 16:36
    
Well, think about $\mathbb{R}$ covering $S^1$, or $\mathbb{C}\setminus \{ 0 \}$ covering itself with the map $z \mapsto z^n$. –  Andy Dec 12 '12 at 17:49
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2 Answers

It follows from your assumptions that $q$ is a 1-sheeted and is a homeomorphism. I'm going to call the map $\pi$ instead of $q$ for the rest of this post.

Assume we have a covering $\pi:X\rightarrow Y$ and $f:Y\rightarrow X$ with $\pi\circ f = Id_X$.

I claim that $f(Y)$ is both open and closed in $X$.

To see it, for any $\hat{p}\in X$, let $p = \pi(\hat{p})$. Choose a neighborhood $U$ around $p$ for which $\pi$ trivializes: $\pi^{-1}(U) = \coprod V_\alpha$ with $\pi|_{V_\alpha}$ a homeomoprhism. and let $V$ be the particular $V_\alpha$ containing $\hat{p}$.

Now, if $\hat{p}\in f(Y)$, then $V\subseteq f(Y)$. This follows from considering the inclusion $i:U\rightarrow Y$. Since both $f|_{U}$ and $\pi^{-1}|_{U}$ are lifts of this inclusion agreeing at $\hat{p}$, they must agree on all of $U$. It follows that $V=\pi^{-1}(U) = f(U)$ as claimed. This shows $f(Y)$ is open.

If, on the other hand $\hat{p}\notin f(Y)$, a very similar argument shows that $V\cap f(Y) = \emptyset$, showing that $f(Y)^c$ is open, that is, that $f(Y)$ is closed.

Putting this together, $f(Y)$ is open and closed. Hence, it is a connected component of $X$. If $X$ itself is connected, this implies $f(Y) = X$ which implies that $\pi$ is a homeomorphism with inverse $f$ so, is in particular, 1 sheeted.

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Haruki, in comments to his own answer above, says s/he thinks there may be a typo in this answer. I don't see one, but perhaps someone else does? –  Jason DeVito Jan 19 '13 at 0:11
    
Do $f|_U$ and $\pi^{-1}|_U$ agree on all of $U$, even if $U$ is not connected? I know about the unique lifting property (see Hatcher, p.62), but it only works if $U$ is connected. –  Stefan Hamcke Feb 18 '13 at 17:14
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@Stefan: I think you're right that if $U$ is disconnected, there could be a problem. On the other hand, I think a decent amount of covering space theory falls apart if one doesn't have nice enough spaces. For example, as you mention, Hatcher requires spaces to be locally path connected (which is enough to save my argument since we can shrink $U$ to be path-connected in this case). I think being locally path connected is also necessary to guarantee the existence of universal covers (it's certainly necessary in the usual proof). But thanks for pointing that out! –  Jason DeVito Feb 18 '13 at 18:15
    
@Stefan: So, in short, I guess I'm assuming the "usual" hypothesis on $Y$. In more detail, I think the only nontrivial property I'm assuming on $Y$ is that it is locally path connected, but there could be something else hidden somewhere. –  Jason DeVito Feb 18 '13 at 18:17
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A connected covering space $f:E\rightarrow X$ admits no section ( global section) unless $f$ is a homeomorphism.

Edit: looking @Andy's post I'm not so sure of what I said now.

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@rschwieb I'm not sure i understand what you are implying? –  Haruki Dec 12 '12 at 18:49
    
Your edit says you were looking at "Andy's post," but now I see you meant that it cast doubt on your solution, not that you were addressing it. Nevermind! –  rschwieb Dec 12 '12 at 18:57
    
@rschwieb yes,it cast doubt on my solution...I am no longer sure that what I said holds...maybe you could shed some light. –  Haruki Dec 12 '12 at 18:59
    
Nope, not familiar enough with the terms :) Good luck! –  rschwieb Dec 12 '12 at 19:00
    
Ok. Maybe someone else could give a more thorough explanation! –  Haruki Dec 12 '12 at 19:39
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