# Strong deformation retract of a singleton implies locally path connected at that point?

Strong deformation retract of singleton $$\{x\}$$: there exists a continuous $$H: X \times I \to X$$ s.t. $$\forall t \in I: H(x,t) = x$$, $$\forall y \in X: H(y,0) = y$$ and $$\forall y \in X: H(y,1) = x$$.

Locally path connected at a point $$p$$: there exists an open neighborhood basis of $$p$$ consisting of path connected sets.

So the question is where $$\{p\}$$ a strong deformation retract implies that $$X$$ is locally path connected at $$p$$.

I haven't been able to come up with any counterexamples: all spaces I have consider which are contractible are locally path connected at all points that are strong deformation retracts.

I have prove the intermediate lemma, which would likely factor into to any proof of the affirmative: $$\{p\}$$ is a strong deformation retract implies that for all open neighborhoods $$U$$ of $$p$$, there exists an open neighborhood $$V \subseteq U$$ of $$p$$ s.t. $$\forall y \in V$$ there exists a path from $$y$$ to $$p$$ lying entirely in $$U$$.

This can be proved by considering $$H^{-1}(U)$$, where $$H$$ is the homotopy described above, noting that $$\{p\} \times I \subseteq H^{-1}(U)$$, so by the tube lemma, there exists an open $$V \subseteq X$$ s.t. $$V \times I \subseteq H^{-1}(U)$$ and $$p \in V$$. Then the homotopy induces a path from $$y \in V$$ to $$p$$ which lies in $$U$$.

On the other hand I have exhibited a TS $$X$$ s.t. there exists a point $$p$$ s.t. for all open neighborhoods $$U$$ of $$p$$, there exists an open neighborhood $$V \subseteq U$$ of $$p$$ s.t. $$\forall y \in V$$ there exists a path from $$y$$ to $$p$$ lying entirely in $$U$$, and yet $$X$$ is not locally path connected at that point. So the lemma alone isn't sufficient. Unfortunately, this space is not a strong deformation retract (I'm fairly certain) to that point $$p$$, so it is not a counterexample. I can give a construction of the example if anyone wants it.

It is true. Your "intermediate lemma" is a well-known alternative characterization of local path connectivity.

Let us show that the following are equivalent:

1. $$X$$ is locally path connected at $$p$$.

2. For all neighborhoods $$U$$ of $$p$$, there exists a neighborhood $$V \subset U$$ of $$p$$ such that for all $$y\in V$$ there exists a path from $$y$$ to $$p$$ lying entirely in $$U$$. [If this is satisfied, $$X$$ is called locally $$0$$-connected at $$p$$ (short: $$LC^0$$ at $$p$$).]

$$1. \Rightarrow 2.$$: Trivial.

$$2. \Rightarrow 1.$$: Let $$U$$ be a neighborhood of $$p$$ and $$C \subset U$$ be the path component of $$p$$ in $$U$$. Choose a neighborhood $$V \subset U$$ of $$p$$ such that for all $$y\in V$$ there exists a path from $$y$$ to $$p$$ lying entirely in $$U$$. Then clearly all $$y \in V$$ are in $$C$$. Thus $$C$$ is neighborhood of $$p$$ and by definition it is path connected.

Note that $$C$$ is not necessarily an open neighborhood of $$p$$. If you understand "neigborhood" as a synomyn for "open neighborhood", then the proof does no longer work. However, a space $$X$$ is locally path connected iff it is $$LC^0$$ (i.e. $$LC^0$$ at all points). In fact, now $$U$$ is open which implies that $$U$$ is also a neighborhood of each $$p'$$ in the path component $$C \subset U$$ of $$p$$ and we see as above that $$p'$$ has a neighborhood $$V \subset U$$ which is contained in $$C$$.

• Good answer. This gives a slightly weaker result than what I asked in the question though (there was some ambiguity since I didn't write open neighborhood in my definition of locally path connected at a point, though that is the standard definition I think): namely that if every singleton is a strong deformation retract then $X$ is locally path connected. The question (which is starting to seem false, though I don't know of a counterexample) is whether strong deformation retract of $\{x\}$ implies locally path connected at $x$. Nov 25 '19 at 2:41
• I do have an example where $X$ is $LC^0$ at $x$ but not locally path connected at $x$. I don't think though it is a strong deformation retract at $x$, though haven't been able to prove it's not. Nov 25 '19 at 2:44
• Defining "locally path connected" via arbitrary neighborhoods or open neigborhoods is not really standardized in the literature, although wikipedia supports your point of view en.wikipedia.org/wiki/Locally_connected_space Anyway, you clarified this in the question, and it remains open. Nov 25 '19 at 9:08

Let X be the union of all lines through (0,0) with rational slope.
Let H be a retract of X to {(0,0)}.
X is path connected and not locally connected.

• But it is locally path connected at $(0,0)$. Nov 25 '19 at 9:06