# If $A$ and $D$ are simply connected subspaces with $A \cap D \neq ø$ then $A \cup D$ is also simply connected.

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.

## 2 Answers

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, $$I_1=\{(x,y)\in C, x\geq 0\}, I_2=\{(x,y)\in C, x\leq 0\}$$.

• But why can we affirm that the circle $C$ = $I_1 \cup I_2$ is not simply connected? Is path connected but what is the path that hasn't an homotopy with the constant path?
– user732763
Jan 17 '20 at 18:59
• @Alvaro: The circle is a counterexample. Jan 17 '20 at 19:01
• math.stackexchange.com/questions/1798923/… Jan 17 '20 at 19:06
• $S^1$ is the circle. Jan 17 '20 at 19:07
• @Alvaro in topology we typically refer to the space $\mathbb{D}^2 = \{(x, y) \in \mathbb{R}^2\ |\ x^2 + y^2 \leq 1\}$ as a "disk" or "ball", and the "circle" is $S^1 = \{(x, y) \in \mathbb{R}^2\ |\ x^2 + y^2 =1\}$ which is the boundary of the disk. Jan 17 '20 at 22:12

To help you visualize T.Aristide answer, think of a simply connected space as something with no hole, that is something you cannot hang on a needle.

If you take a sheet of paper and punch a hole in it, then it is no more simply connected.

But now if you cut this paper along a line that passes through the hole, both sheets obtained have no holes, they have only a curvy edge (half circle) in some area but you cannot hang the pieces of paper on a needle, they fall off.

So despite both half are simply connected, the union is not.