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Does Semilocally Contractible + Has Open Path components => Semilocally simply connecte? |
Semilocally contractible implies both Semilocally simply connected (=semi-locally 1-connected)and Has Open Path components (=semi-locally 0-connected). I.e., the first is because an inclusion map being null-homotopic implies So we could change the current file to any of:
I prefer the last one because the definitions are set up to make this an "unwrapping the definitions" theorem. Semilocally simply connected does not imply Has Open Path components (semi-locally 0-connected), whereas |
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will try to get to it this weekend |
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For the definition, if Then also there would be no real need to mention the bit about open nbhds. What was your thinking for mentioning the open nbhds here? I am thinking it could be useful to make a similar remark for P229 (SLSC). Thoughts? |
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Regarding the proposed theorem in the second bullet in #1745 (comment), I think it can be done later in a separate PR. That way, we already have P239 out there and we can experiment with it. |
@prabau I think I was probably looking at the It does seem a bit trivial though. I'm not sure it's worth mentioning. Maybe if we saw it used in a trait I'd be more supportive of including it. |
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
Suggested in the comment #1672 (comment). There are some references to it in this comment #1672 (comment). We planned to modify weakly locally contractible upon adding this here #1672 (comment).
For this PR:
I changed the reference from Borges to Sakai because Borges's definition is actually slightly different (though equivalent), and this subtlety seems distracting.
This MSE post shows that Semilocally contractible + Has a group topology implies LC. This theorem could be included now or in a future PR.
Note that, unlike SLSC, this property does not satisfy the meta-property "X satisfies this property iff each of its path components does." I.e. a totally path disconnected + semilocally contractible space is discrete, so$\mathbb{Q}$ is not semilocally contractible. SLSC (which is standard terminology) does not include semilocally 0-connected (= has open path components); maybe it's better to read semilocally simply connected as semilocally 1-connected to remember this. On the other hand, semilocally contractible does imply semilocally 0-connected.