Let $\{U_i\}$ be an open covering of the space $X$ having the following properties: (a) There exist a point $x_0$ such that $x_0\in U_i$ for all $i$. (b) Each $U_i$ is simply-connected. (c) If $i\not= j$ then $U_i\cap U_j$ is arcwise connected. Prove that $X$ is simple connected.

Given any loop $f:I\rightarrow X$ based at $x_0$ I have to prove that this is homotopic to the constant map $e(t)=x_0$

Hint: Consider the open cover $\{f^{-1}(U_i)\}$ of the compact metric space space $I$ and make use of the Lebesgue number of this covering.

What I have done: Take $r$ as the Lebesgue number of the open cover $\{f^{-1}(U_i)\}$ and consider a division of $I$ as $[t_0=0,t_1],..., [t_{n-1},1=t_n]$ where $t_{j}-t_{j-1}<r$ for each $j$. Then for each $1\leq j\leq n$ there exist an $i$ such that $[t_j,t_{j-1}]\subset f^{-1}(U_i)$ the $f[t_j,t_{j-1}]\subset U_i$. Therefore we have a finite family $\{U_1,...,U_n\}$ of open subset of the open cover of $X$ given such that $f[t_j,t_{j-1}]\subset U_j$. I don't what to do with at this point

Any advice?



For every $i=0,\dots n$, choose a path $$\gamma_i:\{t_i\}\times I\to U_i\cap U_{i+1}$$ from $\gamma_i(1)=f(t_i)$ to $\gamma_i(0)=x_0$. For $i=0,n$, you can let $\gamma_i$ be the constant path at $x_0$. This gives you a map from the subspace $$I\times\{1,0\}\cup\{t_0,\dots,t_n\}\times I$$ of $I\times I$ to $X$ which on $I\times\{1\}$ coincides with $f$ and on $I\times\{0\}$ coincides with the constant map at $x_0$. On each rectangle $[t_{i-1},t_i]\times I$, we can extend the map from its boundary to the entire rectangle since $U_i$ is simply-connected. Do you see how this gives a homotopy from $f$ to the constant map?

  • $\begingroup$ It was a nice construction. And continue the way I was following! $\endgroup$ – YTS Feb 23 '15 at 14:35

Observe that $I$ is compact and its image will lie in finitely many $U_i$ (as you figured out). Also note that Van Kampens theorem will tell you that the union of those finitely many $U_i$ is generated by the fundamental groups of those $U_i$. So you should arrive quickly at your goal.


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.