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.


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

  • $\begingroup$ 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? $\endgroup$
    – user732763
    Jan 17 '20 at 18:59
  • $\begingroup$ @Alvaro: The circle is a counterexample. $\endgroup$
    – copper.hat
    Jan 17 '20 at 19:01
  • $\begingroup$ math.stackexchange.com/questions/1798923/… $\endgroup$ Jan 17 '20 at 19:06
  • 1
    $\begingroup$ $S^1$ is the circle. $\endgroup$ Jan 17 '20 at 19:07
  • 1
    $\begingroup$ @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. $\endgroup$
    – William
    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.


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