Timeline for Topological vs pro fundamental groups
Current License: CC BY-SA 2.5
9 events
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| Feb 9, 2011 at 16:56 | history | edited | Jeremy Brazas | CC BY-SA 2.5 |
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| Feb 9, 2011 at 14:30 | comment | added | Jeremy Brazas | The definition I give is not an uncommon one. Some might also add the condition $\pi_1(U,y)=1$ for each $y\in U$. It seems appropriate in the setting of "bad" spaces since local path connectivity has little to do with loops being null-homotopic within small neighborhoods. | |
| Feb 9, 2011 at 6:24 | comment | added | Mike Shulman | Wikipedia says that a locally simply connected space is one that admits a basis of simply connected sets, which implies local path-connectedness: en.wikipedia.org/wiki/Locally_simply_connected_space | |
| Feb 7, 2011 at 20:17 | comment | added | Jeremy Brazas | This is what I am familiar with: A space is locally simply connected at a point $x$ if there is a neighborhood base at $x$ consisting of open sets $U$ with $\pi_{1}(U,x)=1$ (forgetting path components of $U$ not containing $x$. The space is locally simply connected if it is so at all of its points. | |
| Feb 7, 2011 at 5:57 | comment | added | Mike Shulman | Also, how can a space be locally simply connected without being locally path-connected? To me "simply connected" implies "path connected," and how can that not be true locally as well? | |
| Feb 7, 2011 at 5:53 | comment | added | Mike Shulman | Well, I'm happy to say that defining $\pi_1$ in terms of maps out of $S^1$ at all is "obviously wrong" in the non--locally-path-connected case. The connection to open subgroups is intriguing because the open subgroups of a profinite group also correspond to its discrete quotients, hence to covering spaces for the profinite fundamental group. Although of course the profunite fundamental group is an honest topological group. | |
| Feb 6, 2011 at 0:43 | comment | added | David Roberts♦ | Ah, that implication in your penultimate paragraph is nice! The open embedding $\langle e \rangle \to \pi_1(X)$ forcing the topology on $\pi_1(X)$ to be discrete - sweet. | |
| Feb 6, 2011 at 0:30 | history | edited | Jeremy Brazas | CC BY-SA 2.5 |
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| Feb 6, 2011 at 0:23 | history | answered | Jeremy Brazas | CC BY-SA 2.5 |