They ask me to decide if it is true or false the next sentence:
- If $A$ and $D$ are simply connected subspaces with $A \cap D \neq ø$ then $A \cup D$ is also simply connected.
Well my answer is:
I think it is true because if $A \cap D \neq ø$ then the space $A \cup D$ is path connected so the first statement of the definition of simply connected is OK, and then to prove that the fundamental group of $A \cup D$ is the trivial one I write that as long as it is path connected if we prove that for one point of $A \cup D$ the fundamental group is the trivial it will be for every point. So if you take any path ($\delta$) that is contain in $A \cup D$ it will be on the class of the constant path ($C_p$) because we don't have any obstacles or missing points so the homotopy define as $H(s,t) = (1-s) \cdot \delta$ + $s \cdot C_p$.
And that's all, I don't know if it is right or not because I see one answer from previous years and it says it is false.
Best Answer
Two halves of a circle are simply connected but not the circle. Another way to say that: a circle can be viewed as the union of two simply connected intervals which intersect in two points. For example take the circle $C$ centered at $0$ of radius $1$ in the plane, $I_1=\{(x,y)\in C, x\geq 0\}, I_2=\{(x,y)\in C, x\leq 0\}$.