You are right... sorry..!
I am really confused about this topic. I am trying to prove something about, but when I go a step forward I find a new complication.
I was trying to avoid to prove directly that something is simply connected... but I guess there are no ways to find rules about...
Actually I am not able to find a counter example.
Intuitively I think that either the intersection of two simply connected is not connected (two or more components) or if it is path-connected then it should be also simply connected.
Does it make sense?
Hi guys...
I was thinking about my questions and another question arises...
if I have two simply connected spaces with path-connected intersection, could I conclude that it is also simply connected?
Intuitively I think so, but I am not able to prove it or to find a counterexample!
sorry... what do you mean by "In the example of the spheres they are convex"?
Do you have suggestions, how to find other examples where between points there are infinitely many geodesics?
I know that in general the intersection of two simply connected spaces is not simply connected.
But for example, if I consider R^2 equipped with the metric associated to the l^1 norm, then any path whose image is the graph of a monotonic function, is a geodesic segment. Therefore there are...
"Do you mean that the subspaces are also geodesic in the sense that any two points can be connected by a geodesic that is also a geodesic in the big manifold?"
Exactly!
I know that in general the intersection of two simply connected spaces is not simply connected. I was wondering if...
A geodesic space is a space where for all two points there exists at least one geodesic joining them.
A geodesic is the shortest path between points, but I am considering a general case where there might be infinite shortest paths between two points.
Hi everybody!
I have a question, if I have A and B simply connected subspaces of a geodesic space X, what can be said about their intersection?
When is it simply connected? Are there rules for this?
I need to prove it in a special case, but I am not able to do it and I was wondering if...