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So I've been working through this book (A Basic Course in Algebraic Topology, by William Massey) in preparation for an Algebraic Topology course I'm going to take soon, and I ran into trouble with this exercise.

Let ${U_i}$ be an open covering of the space X having the following properties:

(a) There exists a point $x_0$ such that $x_0$ ϵ $U_i$ for all i.

(b) Each $U_i$ is simply connected.

(c) If i≠j, then $U_i$ ∩ $U_j$ is arcwise connected.

Prove that X is simply connected.

The book hints at proving that any loop f: I$\to$X (where I = [0,1] ) based at $x_0$ is trivial. I know I should use the fact that {$f^{-1}$($U_i$)} is an open covering of the compact space I, which means that there exists a finite subcovering of I. However, I am not sure how to proceed from there. Any hint at all would be appreciated. Thanks in advance.

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mynameissurge
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    The image of $I$ in $X$ is compact, so is covered by finitely many $U_i$. What can we say about the segment of the loop within a single (simply connected) $U_i$? – ziggurism Oct 10 '14 at 00:13
  • Can such a segment be contracted to a point? – mynameissurge Oct 10 '14 at 00:30
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    simply connected means that every loop can be contracted to a point. How can I turn my segment in $U_i$ into a loop? – ziggurism Oct 10 '14 at 00:32
  • Well, for a segment to be a loop, the endpoints have to be identified with each other. How does this help? – mynameissurge Oct 10 '14 at 00:34
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    How can I turn my segment, which is contained in $U_i$, into a loop contained within $U_i$, based at $x_0$ which is also contained in $U_i$? – ziggurism Oct 10 '14 at 00:36
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    Each $U_i$ is path connected, hence every loop (based at $x_0$) can be written as product of a loops (based at $x_0$) included in $U_i$. since $U_i$ is simply connected those loops are trivial. – Hamou Oct 10 '14 at 00:57
  • Can I choose a path class with initial and terminal point $x_0$ contained in $U_i$ which contains the segment? – mynameissurge Oct 10 '14 at 01:01

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Take a path $\phi\colon I \to X$ starting and ending at $x_0$. From the Lebesgue number lemma (Proof of the Lebesgue number lemma) for the covering $\phi^{-1}(U_i)$ of $I$ we conclude that there exists $n \ne 1$ so that $[\frac{k-1}{n}, \frac{k}{n}]$ is inside some $\phi^{-1}(U_i)$, that is, $\phi([\frac{k-1}{n}, \frac{k}{n}] )\subset U_{i_k}$. For $1< k <n$ , $k$ natural, the point $\phi(\frac{k}{n})$ is in $U_{i_k}\cap U_{i_{k+1}}$, and so is $x_0$. Take a path $\psi_k \colon I \to U_{i_k}\cap U_{i_{k+1}}$ connecting $x_0$ and $\phi(\frac{k}{n})$.

Here are the steps :

Decompose $\phi$ into a composition of $n$ paths $\phi_k$ that are basically the restrictions of $\phi$ to $[\frac{k-1}{n}, \frac{k}{n}]$

Show that

$$\phi_1 \star \ldots \star \phi_n \simeq \phi_1 \star \psi_1^{-1} \star \psi_1 \star \phi_2 \star \psi_2^{-1} \ldots \star \phi_n $$

For each $k$ the path $\psi_{k-1}\star \phi_k \star \psi_k$ is a closed look in $U_{i_k}$ and therefore homotopic to a trivial path.

Piece together to get a homotopy from the loop $\phi$ to the constant loop at $x_0$.

There are some alternate, equivalent ways to define the path composition and the equivalence of paths, some making certain proofs easier -- compare with the path composition in the book by Crowell and Fox, Knot Theory http://www.maths.ed.ac.uk/~aar/papers/crowfox.pdf.

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orangeskid
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