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?
If a covering map has a section, is it a $1$-fold cover?
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1Does it have *only one* section? – 2012-12-12
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2Well, what if $E = X \amalg X$? – 2012-12-12
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0@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. – 2012-12-12
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0@Andy I'm not sure how that makes a difference? – 2012-12-12
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0Well, think about $\mathbb{R}$ covering $S^1$, or $\mathbb{C}\setminus \{ 0 \}$ covering itself with the map $z \mapsto z^n$. – 2012-12-12
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0Addressing the other question by OP in the answers, I would like to suggest the following: $f\colon x \mapsto (x,0)$ and $q\colon(x,y)\mapsto x$ – 2012-12-12
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0For $p$ to have a section, does $p$ have to be a homeomorphism? – 2012-12-12
3 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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0Haruki, 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? – 2013-01-19
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1Do $f|_U$ 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. – 2013-02-18
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1@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! – 2013-02-18
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0@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. – 2013-02-18
Why an answer to a six year old question? Simply because it is an interesting question and the existing answer applies only under the assumption that $X$ is locally connected.
Without any local niceness assumption we shall prove:
Let $q : E \to X$ be a covering map with connected domain $E$. If $p$ has a section $f : X \to E$, then $p$ is a $1$-fold covering (which is the same as a homeomorphism).
The answer is given as a community wiki because the essential idea is contained in the question Section of a covering projection from a connected space which was closed as a duplicate of the present one.
Let us first observe that $1$-fold coverings are nothing else than homeomorphisms. A $1$-fold covering is obviously a bijection. Since all coverings are open maps, we see that $1$-fold coverings are homeomorphisms. The converse is trivial.
To prove that $q$ is a homeomorphism, it suffices to show that $f(X) = E$. Then $f \circ q \circ f = f \circ id_X = f = id_E \circ f$ which implies $f \circ q = id_E$ because $f$ is surjective.
$f(X) = E$ will be proved by showing that $f(X)$ is open and closed in $E$.
Let $y \in E$. There exists an open neigborhood $U$ of $q(y)$ in $X$ which is evenly covered, i.e. we have $q^{-1}(U) = \bigcup_{\alpha \in A} U_\alpha$ with pairwise disjoint open $U_\alpha \subset E$ such that the restrictions $q_\alpha : U_\alpha \to U$ are homeomorphisms.
Let $\alpha(y)$ be the unique index such that $f(q(y)) \in U_{\alpha(y)}$. Since $f$ is continuous, there exists an open neighborhood $U' \subset U$ of $q(y)$ such that $f(U') \subset U_{\alpha(y)}$. Obviously $q_{\alpha(y)}(f(U')) = U'$, hence $f(U') = (q_{\alpha(y)})^{-1}(U')$.
As a subset of $U$ also $U'$ is evenly covered, with decomposition $q^{-1}(U') = \bigcup_{\alpha \in A} U'_\alpha$, where $U'_\alpha = U_\alpha \cap q^{-1}(U') = q_\alpha^{-1}(U')$.
By construction $q^{-1}(U') \cap f(X) = f(U') = U'_{\alpha(y)}$.
If $y \in f(X)$, we have $y \in q^{-1}(U') \cap f(X) = U'_{\alpha(y)}$ which is an open subset of $E$ contained in $f(X)$. This shows that $f(X)$ is open.
If $y \notin f(X)$, we have $y \in q^{-1}(U') \setminus f(X) = \bigcup_{\alpha \in A \setminus \{ \alpha(y)\}} U'_\alpha$ which is an open subset of $E$ not intersecting $f(X)$. This shows that $f(X)$ is closed.
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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0@rschwieb I'm not sure i understand what you are implying? – 2012-12-12
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0Your 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! – 2012-12-12
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0@rschwieb yes,it cast doubt on my solution...I am no longer sure that what I said holds...maybe you could shed some light. – 2012-12-12
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0Nope, not familiar enough with the terms :) Good luck! – 2012-12-12
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0Ok. Maybe someone else could give a more thorough explanation! – 2012-12-12
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0@Jason deVito I think there might be a typo in your solution – 2012-12-12
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0@Haruiki: Sorry, I wasn't pinged because I hadn't yet taken part in the comments under this answer. So, if you try pinging me again, it should work now ;-). Where is the typo? I'd love to fix it (if I can!). – 2013-01-19