[Math] Prove Simply Connected

algebraic-topologyconnectednessfundamental-groupsgeneral-topology

If $X = U \cup V$ with $U,V$ open and simply connected and $U \cap V$ is path connected, why is $X$ simply connected?

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[Math] Is path-connected and locally path-connected equivalent to simply connected

A simply connected set of course (by definition) is path-connected, but on the other side a path-connected and locally path-connected set need not to be simply connected. The easiest example is an open disk $B \subset \mathbb{R}$ around $0$, with an additional point $0'$, and adding to the topology base a base around $0$, with $0$ substituded with $0'$. Figuratively, $0$ is "doubled".

Any continuous homotopy to this space, starting with a path through $0$, cannot contain $0'$, because of contonuity (the preimage of the paths passing through $0'$ and of the paths passing through $0$ would disconnect $[0,1]$, which is bviously connected) so paths through $0$ and through $0'$ are not homotopically equivalent.

General Topology – Is ‘Connected, Simply Connected’ Redundant?

I think some authors require that "simply connected" only be applied to spaces that are already known to be path connected, and then the definition is that a path connected space is simply connected if its fundamental group is trivial (with some basepoint). Here is an article from 1955 which explicitly uses some version of this convention.

The problem is that if a space is not path connected then there is even more ambiguity than usual about what you mean by taking its fundamental group - now it really matters which path component you pick a basepoint in.

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[Math] Is path-connected and locally path-connected equivalent to simply connected

General Topology – Is ‘Connected, Simply Connected’ Redundant?

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