# Simpley Connectedness.

1. Sep 23, 2008

### MathematicalPhysicist

Let X1,X2 be two spaces which are simple connected, and let their intersection be path connected, show that their union is simple connected.

I think I only need to show that their intersection is simply connected, cause then I think that the fundemental group of X1UX2 is isomorphic to to th union of the fundemental groups of X1,X2 and their intersection (correct?).

Well, if we look at a point x in the intersection, then there's a loop in X1, and a loop in X2, if we look at the path that is common to both loops in the intersection then it's also a loop (if we choose the direction of both loops to be counterclockwise or clockwise) in the intersection and its homotopic to the constant loop, cause this loop is both in X1 and X2, that way we get that the intersection is simply connected.

Is this way off?

2. Sep 25, 2008

### MathematicalPhysicist

X1 and X2 are open in the set X1UX2.

3. Sep 25, 2008

### mathwonk

well think of the example of the upper and lower hemispheres of a sphere. your statement on simple connectivity of the intersection, a circle, is false but the desired statement on the simple connectivity of the union, the sphere, is still true.

another example however is the upper and lower halves of a circle, both s.c. but the intersection is disconnected and the union not s.c.

4. Sep 25, 2008

### tim_lou

doesn't that sound very similar... to a simple version of Van-Kampen's theorem?? I think you need their intersection be simply connected.

5. Sep 26, 2008

### MathematicalPhysicist

mathwonk, why does the intersection of the two halves of the circle are disconnected?
this intersection is the diameter along the horizontal axis, if this interval was broken into more than one part then it would be disconnected (separated), but it's not.

6. Sep 26, 2008

### MathematicalPhysicist

So if it's simply connected, how would you go around it?

7. Sep 26, 2008

### yenchin

The intersection are two points, not the diameter. The circle S^1; not the disc, so there is no interior.

8. Sep 26, 2008

### Doodle Bob

lqg, you're paying too much attention to the wrong part of tim_lou's remark. Check out Van Kampen's Theorem.

9. Sep 26, 2008

### MathematicalPhysicist

So your'e telling me this question is just another implication of a known theorem?!

I'm starting hating university courses.
(-:

10. Sep 26, 2008

### MathematicalPhysicist

Ok Iv'e, looked for Van Kmepen-siefret theorem, and if the intersection is simply connected then $$\pi_1(X1,x_0)*\pi_1(X2,x_0)$$ is isomorphic to $$\pi_1(X1UX2,x_0)$$ where * is the product of functions.
from here it's really obvious.

jesus christ, we just haven't learnt anything in this course, can you prove the problem here without this theorem?