4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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) $U_i$? $\endgroup$ – ziggurism Oct 10 '14 at 0:13
  • $\begingroup$ Can such a segment be contracted to a point? $\endgroup$ – mynameissurge Oct 10 '14 at 0:30
  • 1
    $\begingroup$ simply connected means that every loop can be contracted to a point. How can I turn my segment in $U_i$ into a loop? $\endgroup$ – ziggurism Oct 10 '14 at 0:32
  • 1
    $\begingroup$ 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 $U_i$? $\endgroup$ – ziggurism Oct 10 '14 at 0:36
  • 1
    $\begingroup$ 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. $\endgroup$ – Hamou Oct 10 '14 at 0:57
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.