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I'm having trouble solving the following exercise in Hatcher's Algebraic Topology(1.3 #5):

"Let $X$ be the subspace of $R^2$ consisting of the four sides of the square $[0,1] \times [0,1]$ together with the segments of the vertical lines $x=\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots$ inside the square. Show that for every covering space $\tilde{X}\to X$ there is some neighbourhood of the left edge of $X$ that lifts homeomorphically to $\tilde{X}$. Deduce that $X$ has no simply-connected cover."

I tried piecing together open sets in $\tilde{X}$ which were homeomorphic to open square shaped $\epsilon$-neighbourhoods of points in the left edge of $X$, but I couldn't glue them together coherently.

My second idea was to use the path lifting property to lift the left edge, but I couldn't extend it to a lift of an open neighbourhood of the left edge.

Edit: Some more detail on my first attempt:

Suppose I have two open squares $U_1$ and $U_2$ in $X$ (containing the left edge) with nonempty intersection, and that have preimages $\coprod_i U_{1,i}$ and $\coprod_j U_{2,j}$ with each piece of that mapping homeomorphically onto $U_1$ and $U_2$ respectively.

Suppose I've chosen a specific $U_{1,i}$, and I want to choose a specific $U_{2,j}$ so that their union maps homeomorphically to $U_1$ union $U_2$. Such a choice can be specified by a point in the chosen $U_{1,i}$ that maps into $U_2$, but for each such point, there could be a different choice of $j$.

My intution tells me that to make this choice, one needs to appeal to the bad local nature of the left edge of $X$, (so that you're certain to get some of the vertical lines inside of the desired open set) but I'm not sure how to proceed with that.

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I think this should come directly from the definition of a covering space, where you've got that the projection map has to be a local homeomorphism. And probably you'll need to use that the interval is compact, to extract a finite subcover of a cover whose open sets all lift homeomorphically. – Aaron Mazel-Gee Feb 6 '11 at 1:30
why can't you glue the open squares coherently? I think this follows from the compactness of left edge. – Soarer Feb 6 '11 at 1:39
@Soarer, Aaron: I've added some detail on why I couldn't glue them. – alephzero314 Feb 6 '11 at 2:49
Do you mean that some part of the left edge is contained in $U_1$ and $U_2$? I don't think you can assume they contain all of the left edge. As said above I think you want to start with an open cover of the left edge and show you can paste them together using a compactness argument. – Joe Johnson 126 Feb 8 '11 at 13:29

The trick in gluing here is, as you correctly pointed out, to fix a point in $U_{1,i}$. Fix a point in $U_{1,i}$ such that its image under the covering map goes to $U_1\cap U_2$. Thus it specifies a point which belongs to $U_{2,j}$ for some $j$. And hence we can get a common neighbourhood.
The bad local nature is used in the second step. As a neighbourhood of the left edge can be lifted homeomorphically, we can clearly see that there are non trivial loops in the $\tilde{X}$.

Hope the explanation is clear enough.

EDIT: This is my answer to the comment made by joriki. I couldn't write the whole thing as a comment hence I am writing it here.

I don't think one will be able to prove that the circle lifts homeomorphically to an open interval using the idea I suggested. The issue that will occur is the following: consider a cover of circle with two open sets $U_1$,$U_2$ such that both $U_1$ and $U_2$ can be lifted homeomorphically and choose a $U_{1,i}$ and $U_{2,j}$ as suggested. What happens in this case is that the projection map will not be injective on the union of $U_{1,i}$ and $U_{2,j}$, as $U_{1,i}\cap U_{2,j}$ will have two components. As soon as we fix a pre-image for a point in one intersection the preimage of $U_2$ is determined. Now $\pi(U_{2,j}-U_{1,i})\cap U_1$ will not be empty(Here $\pi$ is the projection map). In the general case when we take $n$ open sets what happens is that $\pi(U_{n,i_n}-\cup_{1}^{n-1}U_{k,i_k})\cap U_1$ will be non empty and hence the projection map will not be injective. The thing is that, any proper subset of the circle can be lifted homeomorphically can be proved by the arguement i have suggested. The issue occurs precisely because the circle closes on itself.

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It seems that this would prove that the circle lifts homeomorphically to an open interval of the line? (Take the circle covered by the line, consider it as a "rectangle" embdedded in the unit square with one edge playing the role of the left edge of the comb space, and choose neighbourhoods of points on that edge that also contain the corresponding parts of the opposite edge.) So it seems we do have to use the "bad local nature" (I like that :-) in this step. – joriki Feb 9 '11 at 12:49
up vote 0 down vote accepted

I believe that I found an answer to my problem. The earlier approach of playing with open covers didn't pan out because I couldn't extend local injectivity of $p$ on a bunch of open sets to injectivity on their union.

But I found the following theorem, which solves the problem quite neatly. I found it in [1], along with a sketch proof. (But I'll prove it fully):

Theorem: Let $E$ be any space and $F$ be a metric space. Let $K\subset E$ be compact, $f:E\to F$ be continuous, locally injective near every point $p\in K$, and (globally) injective on $K$. Then $f$ is also injective on some open neighbourhood $N$ of $K$.

Proof: Let $g:E\times E\to R$ be the function defined by $g(x,x')=d(f(x),f(x'))$. Around each point of the form $(k,k)\in K\times K$, there is an open product neighbourhood $V_k \times V_k$ such that $(V_k\times V_k)\cap g^{-1}(0)^c \subset \{ (x,x)\in E\times E \}$ (by local injectivity). Let $V=\cup_k V_k$. Then the compact set $K\times K$ is contained in the open set $g^{-1}(0)^c \cup V$. By the generalized tube lemma, there are open sets $W,W'$ of $E$ so that $K\times K\subset W\times W'\subset g^{-1}(0)^c \cup V$. Then let $N=W\cap W'$. $\square$

Applying this theorem to $p:\tilde{X} \to X$, with $K$ being a lift of the left edge of $X$, gives an open neighbourhood of that lift which maps homeomorphically to an open neighbourhood of the left edge of $X$. Then we use the bad local nature of $X$ to find a loop in $X$ which lifts to a loop in $\tilde{X}$ which is not contractible. (Compactness is used again to find an open set which contains the entirety of a vertical line)

[1] Wagner,M., "Existence of schlicht integral manifolds for ordinary differential equations", Archiv der Mathematik, Volume 61, Number 6, 529-542 (1993)

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