Prove Simply Connected

Question

If $X = U \cup V$ with $U,V$ open and simply connected and $U \cap V$ is path connected, why is $X$ simply connected?

Answer 1

Hint: Apply the van Kampen theorem.

Answer 2

As Ayman Hourieh said, one can use van Kampen's theorem.

But in the present case, it might worth it to prove it by hands to really understand what is going on. Such a direct proof goes as follow (it is kind of the proof of van Kampen's theorem, but in this really simple case) : given a loop $\gamma \colon [0,1] \to X$,

• choose a base point $x_0$ in $U \cap V$,
• pull the cover $\{U,V\}$ back to $[0,1]$ and apply Lebesgue's lemma to get a subdivision $0=i_0 < i_1 < \dots < i_n = 1$ such that for all $k$, $\gamma \!\!\restriction\!\! [i_k,i_{k+1}]$ has its image entirely either in $U$ or $V$,
• for each portion $\gamma \!\!\restriction\!\! [i_k,i_{k+1}]$ of the loop $\gamma$, construct a loop with base point $x_0$ contained either in $U$ or $V$,
• apply simple connectedness in $U$ and in $V$ to get a homotopy (defined piecewise) from $\gamma$ to the constant loop with value $x_0$.