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Aug 7, 2022 at 10:13 vote accept Joel Springer
Aug 5, 2022 at 11:04 comment added Joel Springer @WlodAA my question was about contractible spaces not locally contractible
Aug 5, 2022 at 9:02 comment added Wlod AA You need to provide the relevant definitions, including the local contractability. (X is loc. contr. at p in X <=:=> for every nbhd U of p there is a nbhd V of p contained in U such that (V p) admits a contraction in (U p), i.e a homotopy on (V p) that connect the identical map of (V p) to (U p) with the constant into p, inside U. ### It's tiring to discuss math without explicit definitions.
Aug 5, 2022 at 7:15 comment added Joel Springer Thank you @HenrikRüping - could you please elaborate on how you know this example is not homologically locally connected?
Aug 5, 2022 at 7:03 comment added HenrikRüping en.wikipedia.org/wiki/Cone_(topology). Locally (away from the special point) the Cone looks like the original space crossed with an interval, so taking the cone preserves weird local properties. But globally the cone is contractible (move things to the special point).,
Aug 5, 2022 at 6:58 comment added Joel Springer @HenrikRüping What do you mean by the cone over in this case?
Aug 5, 2022 at 6:47 answer added Sam Nead timeline score: 2
Aug 5, 2022 at 6:38 comment added HenrikRüping No. Take the cone over $\{(x,sin(1/x))\mid x>0\}\cup \{(0,t)\mid 0\le t\le 1\}$.
Aug 5, 2022 at 6:21 comment added Joel Springer @WlodAA Could you please elaborate on why (for a contractible space) there exists a $V$ where the inclusion in $U$ is homotopic to constant (I think you are saying this is true for any $x \in V \subset U$ right?)
Aug 5, 2022 at 6:08 comment added Wlod AA The local contractibleness does not talk about homotopy equivalences but about contractions of sub-neighborhoods inside the neighborhoods.
Aug 5, 2022 at 6:00 comment added Wlod AA Now, it's instant. The inclusions of the sub-neighborhoods into the neighborhoods are homotopic to constant hence they are homologically trivial (i.e. the zero-homomorphisms).
Aug 5, 2022 at 5:52 history edited Joel Springer CC BY-SA 4.0
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Aug 5, 2022 at 4:08 history edited Joel Springer CC BY-SA 4.0
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Aug 5, 2022 at 4:00 comment added Jeremy Brazas Perhaps you mean something else for the homological property? As stated, you can just take $V=U$ for given $U$ so that every space satisfies the condition.
Aug 5, 2022 at 3:24 history edited Joel Springer CC BY-SA 4.0
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Aug 5, 2022 at 2:45 history asked Joel Springer CC BY-SA 4.0