Loop space of semi-locally simply connected is locally path-connected

I came across the following statement (link).

Let $$X$$ be a locally path-connected and semi-locally simply connected topological space. Denote by $$\Omega$$ the loop space with basepoint $$x\in X$$, endowed with the compact-open topology. Then $$\Omega$$ is locally path-connected.

The page linked above does not provide any proof. Could anyone provide some reference, or give some hints about how to prove this fact (if it helps, for my purposes I can assume $$X$$ is also simply connected)?

To clarify:

• locally path-connected means that $$X$$ has a basis of path-connected subsets;
• semi-locally simply connected means that $$X$$ has a basis of subsets $$\{U\}$$ such that the inclusions $$\pi_1(U)\to\pi_1(X)$$ are the zero maps (see here for details);
• the loop space with basepoint $$x$$ is the set of continuous functions $$f:[0,1]\to X$$ such that $$f(0)=f(1)=x$$.

It is false.

Let $$X$$ be the cone on the Hawaiian earring $$H = \bigcup_{n=1}^\infty S_n \subset \mathbb R ^2$$, where $$S_n$$ is the circle around $$(0,1/n)$$ with radius $$1/n$$. We have $$S_n \cap S_m = \{(0,0)\}$$ for $$n \ne m$$. We may write $$X = \{ t(0,0,1) + (1-t)(x,y,0) \mid t \in I, (x,y) \in H \} \subset \mathbb R ^3$$. The cone point is $$(0,0,1)$$ and $$X$$ inherits a metric $$d$$ from $$\mathbb R ^3$$. Let $$x_0 = (0,0,0) \in X$$ the basepoint of $$X$$.

$$X$$ is contractible, hence semi-locally simply connected, but it is not locally simply connected locally simply-connected vs. semilocally simply-connected. The compact-open topology on $$\Omega X$$ agrees with the metric topology induced by the $$\sup$$-metric $$d_\infty(\ell,\ell') = \sup \{d(\ell(t),\ell'(t)) \mid t \in I \}$$.

Assume that $$\Omega X$$ is locally path connected.

Consider the constant loop $$c(t) \equiv x_0$$. Let $$W_r = \{ \ell \in X \mid d_\infty(c,\ell) < r \}$$. We find an open path connected neighborhhod $$W'$$ of $$c$$ such that $$W'\subset W_1$$ and $$r > 0$$ such that $$W_r \subset W'$$. Let $$\ell_n$$ be the loop in $$X$$ parametrizing the circle $$S_n \times \{ 0 \} \subset X$$. Take $$n$$ such that $$1/2n < r$$. Then $$\ell_n \in W_r$$. Choose a path $$u : I \to W_1$$ such that $$u_0 = c, u_1 = \ell_n$$. We have $$d_\infty(c,u(t)) < 1$$, hence the loop $$u(t)$$ does not go through $$\{(0,0,1)\}$$. The path $$u$$ yields a homotopy $$u' : c \simeq \ell_n$$ such that all $$u'_t = u(t)$$. By construction $$u'$$ is a homotopy in $$X' = X \setminus \{(0,0,1)\}$$. There is a retraction $$d : X' \to H$$, hence we get $$dc \simeq d\ell_n$$. But this is not true which shows that $$\Omega X$$ is not locally path connceted.

Remark:

The link in your question says: In general, if $$X$$ is locally $$n$$-connected, $$\Omega X$$ is locally $$(n−1)$$-connected. This seems more plausible, although I do not know a proof.

• Thank you for your answer. Although I have to say that the page I linked actually states "Let the space $X$ be locally 0-connected and semi-locally 1-connected [...]. The loop space $\Omega X$ for any basepoint is locally path connected", which (as your answer shows) is plainly false (unless I have missed some other hypotesis). Commented Apr 24, 2019 at 10:05