# Simply connected stuff

1. May 20, 2007

### jostpuur

I know that just because $$A\subset X$$ and $$B\subset X$$ are simply connected in some metric space X, the union $$A \cup B$$ is not neccecarely simply connected. However, the does path connectedness of $$A\cap B$$ suffice for the union to become simply connected? If it does, is there easy proof?

Edit:

I don't change the original question, because it could make first replies seem strange, but anyway, the question that I'm now interested (after the first replies), is that is the claim true if we assume the intersection to be nonempty, and both A and B to be open.

Last edited: May 20, 2007
2. May 20, 2007

### Office_Shredder

Staff Emeritus
Well, let's imagine A is a torus, and B is a part of the torus. A intersect B is path connected, but A union B isn't simply connected.

I'm not that up on simply connected sets, but I'm pretty sure that's right?

3. May 20, 2007

### jostpuur

I meant that we also assume A and B both to be simply connected alone.

4. May 20, 2007

### Moo Of Doom

A isn't simply connected if it's a torus, so it doesn't meet the hypothesis that both A and B are simply connected.

Is the empty set considered path connected? If so, we may simply choose some A and B which are some distance apart, and thus their union isn't even path connected.

If the empty set doesn't count (or you require their intersection to be non-empty), then consider set A to be a "C" shape (annulus with a slice missing), and B to be the missing slice, plus a bit more, so there is an overlap on one side but not on the other. Then their intersection is path connected, but their union (the annulus) is not simply connected.

5. May 20, 2007

### jostpuur

Okey, that was a counter example, but I naturally respond by toughening the assumtions

What if A and B are both open (or both closed)? And let us assume that the intersection is nonempty.

I'm asking this because it looks like these assumtions should imply simply connectedness of the union, but I'm not sure how.

6. May 20, 2007

### Office_Shredder

Staff Emeritus
Ok, right, that makes sense of course

Imagine we're in $$R^2$$ then... a pair of rectangles that overlaps at a square is simply connected, right?

Something like:

http://pbskids.org/sesame/coloring/images/l_grover.gif

The L in that picture (it was the first L I could find, shut up....)

So take one of those with the length of each protrusion some length x, and the overlap a length y. Then take another such set, with one protrusion shortened by a length y, and lie them on top of each other so they overlap at a square and form a larger square with a square shaped hole in the middle. The intersection is a square, the union isn't simply connected, and each set is simply connected, right?

Last edited: May 20, 2007
7. May 20, 2007

### Moo Of Doom

If I get your picture, then either their intersection will not be connected (have some points in either corner), or there will be a tiny rip in the square with a hole in the middle, so that it's like a c-shape again, only with nearly touching ends. Remember that A and B both must be open, so they can't be touching without intersecting.

I think these (new) conditions are now sufficient. Time to start thinking of a proof. :)

8. May 20, 2007

### AKG

Let X be the unit circle, let A be the restriction of X to the upper half plane, and let B = X\A. The intersection of A and B is empty, hence path connected trivially, and each A and B are simply connected, but their union is not. It's easy to modify B slightly so that it doesn't intersect B trivially but such that all the above conditions hold.

9. May 20, 2007

### Moo Of Doom

Yes, but B is not open (or A isn't, depending on if we take the closed or open upper half-plane). The new conditions require both sets to be open. Your counterexample has already been brought up.

10. May 21, 2007

### matt grime

Given the extra, new, assumptions, I think it is true, and can be obtained by considering homotopies of loops. The point is if we have any loop in AuB, it can be split into sections lying wholly in AnB, A\B and B\A, and the hypotheses now make it possible to deform this to the trivial loop, though I confess I haven't thought this through rigorously.

11. May 21, 2007

### mathwonk

i suggest looking at the basic theorem in fundamental groups, but i forget the name, seifert / van kampen?

12. May 21, 2007

### mathwonk

ok i looked it up in hatchers book available online for free, and it says if a space is the union of two path connected open sets and the intersection is path connected, and if the base point lies in the intersection, then the fundamental group of the union is a quotient of the free froduct of the fundamental groups of the twoorigianl sets.

so in particular a [quotient of the] free product of two trivial groups is trivial.

does this do it for you? if not, matts description may be more clear. i.e. just visualize shrinking loops.

Last edited: May 21, 2007
13. May 21, 2007

### jostpuur

So it is then...

I've visualized incorrect claims also, so wasn't sure.

14. May 22, 2007

### mathwonk

see if you can believe it. try to visualize it.