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Spanier mentions that locally contractible implies homologically locally connected but I'm wondering whether contractible implies homologically locally connected?

Definition of homologically locally connected: "A space $X$ is said to be homologically locally connected in dimension $n$ if for every $x \in X$ and neighborhood $U$ of $x$ there exists a neighborhood $V$ of $x$ in $U$ such that $\tilde{H}_q(V) \to \tilde{H}_q(U)$ is trivial for $q \leq n$. It is said to be homologically locally connected if it is homologically locally connected in dimension $n$ for all $n$."

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  • $\begingroup$ 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. $\endgroup$
    – Jeremy Brazas
    Commented Aug 5, 2022 at 4:00
  • $\begingroup$ 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). $\endgroup$
    – Wlod AA
    Commented Aug 5, 2022 at 6:00
  • $\begingroup$ The local contractibleness does not talk about homotopy equivalences but about contractions of sub-neighborhoods inside the neighborhoods. $\endgroup$
    – Wlod AA
    Commented Aug 5, 2022 at 6:08
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    $\begingroup$ No. Take the cone over $\{(x,sin(1/x))\mid x>0\}\cup \{(0,t)\mid 0\le t\le 1\}$. $\endgroup$
    – HenrikRüping
    Commented Aug 5, 2022 at 6:38
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    $\begingroup$ 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)., $\endgroup$
    – HenrikRüping
    Commented Aug 5, 2022 at 7:03

1 Answer 1

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Recall that $E$, the earring space, is the union in the plane of a countable collection of shrinking circles, all tangent to the $y$ axis at the origin. To obtain the desired counterexample, form the cone on $E$.

Edit: My example just above is very similar to that of Henrik Rüping (in the comments) and works for much the same reason (also mentioned by Henrik Rüping in the comments). Seeing the further questions asked by the original poster prompted me to give the example below.

Let $S$ be the following subset of the plane:

$$S = \{(0, 0)\} \cup \{ (1/n, 0) \mid \mbox{$n > 0$ a natural number}\}$$

Let $A = (0, 1)$ be the given point in the plane, called the apex. Let $I_n$ be the closed line segment in the plane connecting the apex $A$ to the point $(1/n, 0)$. Finally, let $I_\infty$ be the closed line segment connecting the apex to the origin $(0, 0)$. We define the cone $C(S)$ to be the set

$$C(S) = I_\infty \cup \left( \cup_{n > 0} I_n \right)$$

equipped with the subspace topology. The space $C(S)$ is sometimes called the broom space. (I think it looks a bit more like a "brush" than a broom, but anyway.)

We make two claims.

Claim 1: Straight-line homotopy gives a deformation retraction of the broom $C(S)$ to its apex $A$.

Thus $C(S)$ is contractible.

Claim 2: The broom $C(S)$ is not (homologically) locally connected at the origin $(0, 0)$.

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  • $\begingroup$ thanks - could you please elaborate on how you know this example is not homologically locally connected? $\endgroup$
    – Joel Springer
    Commented Aug 5, 2022 at 7:27
  • $\begingroup$ (The system doesn't like me to post more comments under the OP hence I am here). ### The posts and comments on the Internet, and in particular on MO, tend to hang in the thin air, without clearly addressing the consecutive related issue. Thus, it is very hard or practically impossible to follow such a Facebook or social media style of discussion. ### The posts and the comments should be as much as possible self-contained or at least there should be very clear references. ### For instance, a long enough quoted string from the respective earlier comment or post can serve as a reference. $\endgroup$
    – Wlod AA
    Commented Aug 5, 2022 at 8:56
  • $\begingroup$ @WlodAA is there another website that you would recommend over this one? $\endgroup$
    – Joel Springer
    Commented Aug 5, 2022 at 11:05
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    $\begingroup$ @JoelSpringer - I've added another example which may be more useful to you. $\endgroup$
    – Sam Nead
    Commented Aug 5, 2022 at 11:22

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