2
$\begingroup$

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 $f(0)=f(1)=x$.
$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Delfad0r Apr 24 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.