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

Question

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.

Answer 1

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 $C$.

Answer 2

There are two different notions of local path connectedness at a point $x\in X$:

• locally path connected at $x$ (if $x$ has a nbhd base of path connected open sets),
• path connected im kleinen at $x$ (if $x$ has a nbhd base of path connected sets).

See this wikipedia page for details. Section 2 in this article has a nice diagram that summarizes the implications:

Now you (and the other answer) have already shown:

If the singleton $\{p\}$ is a strong deformation retract of $X$, then $X$ is path connected im kleinen at $p$.

And the question is whether $X$ is in fact locally path connected at $p$ in that case. The answer in general is no. For an example consider the "infinite broom" $X$ as in this question:

As shown in the linked question, the space $X$ is not locally path connected (it is not even locally connected) at the apex point $p$, but it is path connected im kleinen at $p$.

To show that $\{p\}$ is a strong deformation retract of $X$, it helps to work in a space $Y$, homeomorphic to $X$, consisting of a infinite number of "brooms" side by side to the left, with all brooms of the same size, plus an extra point $p$ at $-\infty$ that compactifies it. Now a strong deformation retraction to $p$ can be built in two steps. First leave $p$ fixed and start moving all other points at the same horizontal speed to the right. That is, each point on one of the slanted lines moves down its line to the right until it reaches the horizontal line, after which is keeps moving to the right on the horizontal line. And each point on the horizontal also moves right, and all points move in unison so that points on a same vertical remain on a same vertical as they move. And if any points reaches the right endpoint $a_1$, it stays there. After a finite amount of time, all points will have moved to the horizontal line. The horizontal part of $X$ (including the point $p$) is homeomorphic to a closed line segment. So in the second phase we can just contract the horizontal part to the point $p$ while leaving $p$ fixed. That finishes the strong retraction to $p$.

(This homotopy uses similar ideas as the one in here.)

Answer 3

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.