# A criterion to prove that a topological space is simple connected.

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

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?
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.