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I am working in this exercise:

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

I just don't know why the argument here and here about pasting the homeomorphisms works.

The argument goes like this. We start by taking a finite cover of the left edge composed of evenly covered open sets. Next we take, from the open cover, an evenly covered neighborhood $U_1$ of $0$ and some $V_1$ in the inverse image such that $V_1\cong U_1$ by the covering map $p$.

Now we take a second open set $U_2$ in the cover which intersects both $U_1$ and its complement. We take some point $e_1$ in $V_1$ such that $p(e_1)\in U_1\cap U_2$ and then, observing that $$p^{-1}(U_2)=\bigsqcup_{i\in I}W_i, W_i\cong_p U_2$$ we take $V_2=W_i$ such that the image of $e_1$ is in $U_1\cap U_2$.

Then $p|_{V_j},j=1,2$ is a homeomorphism, which can be glued to a surjective continuous function $p:V_1\cup V_2\to U_1\cup U_2$. To see it's an homeomorphism it's enough to show it's 1-1.

If I suppose $p(x)=p(y)$ with $x\in U_1,y\in U_2$ then $p(x)\in V_1\cap V_2$. But I don't see how to arrive to a contradiction here, or to $x=y$.

Nell
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1 Answers1

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I was once stuck here, same as you (were). Maybe it's about the perspective. We do say "evenly covered neighborhood of $x$", but in fact this property of "being evenly covered" is not subject to the choice of $x$.

Following your notation, $A:=U_1\cap U_2$ is evenly covered, being the intersection of two evenly covered open sets. Now that we have $V_1\subset \tilde{X}$, a homeomorphic copy of $U_1$, there is some $\tilde{A}\subset V_1$ such that $p|_{\tilde{A}}:\tilde{A}\to A$ is a homeomorphism. Then there is a homeomorphic copy $V_2\subset \tilde{X}$ of $U_2$ that contains $\tilde{A}$, so $A\subset p(V_1\cap V_2)$. Moreover, $p(V_1\cap V_2)\subset p(V_1)\cap p(V_2)=U_1\cap U_2=A$, so $p|_{V_1\cap V_2}$ is a homeomorphism to $A$. This implies we can define the inverse of $p|_{V_1\cup V_2}$ by simply combining $p|_{V_1}^{-1}$ and $p|_{V_2}^{-1}$, which shows $p|_{V_1\cup V_2}$ is a homeomorphism.

Juggler
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