$X=A\cup B$ be an open cover of $X$. If $X,A,B$ are simply connected , then $A\cap B$ path-connected?

Question

I'm trying to prove or find counterexample to the following :

Let $X=A\cup B$ be an open cover of $X$. Assume that $X,A,B$ are simply connected , then $A\cap B$ must be path-connected.

I tried a proof by contradiction : Assume that $A\cap B$ not path-connected , and pick loop which start at one of the connected components, say $C_1$ , travelling through all the other connected components , then returning to $C_1$.

I think the fact this loop is nullhomotopic will imply that A∩B is connected, but I haven't found such a proof.

(I'm not sure if that true)

Have you studied homology yet? – Robert Bell Apr 21 '18 at 11:16 — Robert Bell, Apr 21 '18 at 11:16
Then Mayer-Vietoris works. – Robert Bell Apr 21 '18 at 12:00 — Robert Bell, Apr 21 '18 at 12:00
@RobertBell can you elaborate please ? – Bar B Apr 21 '18 at 12:01 — Bar B, Apr 21 '18 at 12:01

score 3 · Answer 1 · answered Apr 21 '18 at 12:10

You can apply the Mayer-Vietoris sequence:

$$H_1(X)\to \widetilde{H}_0(A \cap B) \to \widetilde{H}_0(A) \oplus \widetilde{H}_0(B)$$

Since $X$ is simply connected, $H_1(X) = 0$. Since $A$ and $B$ are simply connected, they are path connected and so their reduced 0-th homology groups are zero. Therefore, $\widetilde{H}_0(A \cap B) = 0$ and so $A \cap B$ is path connected. Here I am using singular homology.

Mohith Nagaraju · Answer 2 · 2022-10-02T07:53:39.483

1

Here is an interesting example to think about that doesn't prove or disprove the question.

Let $X$ be the quasi-circle shown in the figure, a closed subspace of $\mathbb R^2$ consisting of a portion of the graph of $y = sin(1/x)$, the segment $[−1, 1]$ in the $y$ axis, and an arc connecting these two pieces.

The important point about this example is that $X$ is simply connected even though it somewhat resembles a circle.

Now if you pick $A$ to be an open piece in the top part and $B$ to be an open piece in the bottom part, we get $B,X$ is simply connected, and $A$ is almost simply connected, i.e. all loops are null-homotopic ($A$ fails to be simply connected because it isn't path-connected). Lastly observe that $A\cap B$ is not path-connected!

Edit: $A$ is the intersection of the pink region with the quasi-circle and $B$ is the intersection of the purple region with the quasi-circle.

edited Oct 02 '22 at 07:53

answered Sep 30 '22 at 13:48

Mohith Nagaraju

768

Define a path connecting the point $(0,1) to a nearby point of the sine wave. – coffeemath Sep 30 '22 at 13:54
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Given $A \cup B = X$, which set contains the $(0,0)$ point, and is it really simply connected? – aschepler Sep 30 '22 at 13:55
I think more detail is needed as to what $A$ and $B$ are, I am confused. – hunter Sep 30 '22 at 14:12
The topologist's sine curve, which is what your example uses, and joins with an arc, is the standard example of a connected part of the plane which is not also path connnected. In particular it is not simply connected. – coffeemath Sep 30 '22 at 14:30
@coffeemath consider the path from (0,1) to (0,0) and the travels along the nice arc counterclockwise. – Mohith Nagaraju Oct 02 '22 at 06:58
@aschepler, it is indeed simply connected. See: https://math.stackexchange.com/questions/3705688/to-show-that-warsaw-circle-is-simply-connected – Mohith Nagaraju Oct 02 '22 at 07:07
@aschepler see the diagrams – Mohith Nagaraju Oct 02 '22 at 07:12
The Warsaw circle is indeed simply connected, but the pink part is not. – Paul Frost Oct 02 '22 at 07:35

$X=A\cup B$ be an open cover of $X$. If $X,A,B$ are simply connected , then $A\cap B$ path-connected?

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