Is $\Omega \tilde X \simeq \Omega_0 X$?

Question

Let $\tilde X\to X$ be a universal covering of a based space $X$, with a chosen basepoint. Is $\Omega \tilde X \simeq \Omega_0 X$? Here $\Omega$ denotes the loop space, $\Omega_0$ denotes the connected component of the trivial loop and $\simeq$ denotes homotopy equivalence.

I've realized how little I understand about the topology of loop spaces, and I have found it difficult to deal with it (as I don't really "get" the compact-open topology) and I haven't really found sources on the subject...

The context is algebraic $K$-theory, but I don't think it really matters much here.

Answer 1

The two are homeomorphic under the projection $p_*:\Omega\tilde{X}\to\Omega_0 X$ induced by the covering map $p:\tilde{X}\to X$.

$p_*:\Omega\tilde{X}\to\Omega_0 X$ is bijective. This follows from the general theory of covering spaces (unique path lifting, unique homotopy lifting, etc$\dots{}$) and the fact a path between two based loops in $\Omega(X)$ (equipped with the compact open topology) is the same as a based homotopy, so that $\Omega_0(X)$ is precisely the set of all nullhomotopic loops. This is because the unit interval is locally compact and Hausdorff, see the exponential law.

$p_*$ is continuous. if $[K,O]=\lbrace c:[0,1]\to X\mid c(K)\subset O\rbrace$ is a subbasic open neighborhood of the compact open topology on $\mathrm{Map}([0,1],X)$, where $K\subset [0,1]$ is a compact subset and $O\subset X$ is open, then $$(p_*) ^{-1}([K,O])=[K,p^{-1}(O)]$$ is a subbasic open neighborhood of the compact open topology on $\mathrm{Map}([0,1],\tilde{X})$ since $p^{-1}(O)$ is open by continuity of $p$. Since $p_*$ maps $\Omega(\tilde{X})\subset\mathrm{Map}([0,1],\tilde{X})$ to $\Omega_0(X)\subset\mathrm{Map}([0,1],X)$, it follows that the map you are considering is continuous as a restriction of a continuous map. (The exact same argument shows that for any three spaces $X,Y_1,Y_2$ and continuous map $f:Y_1\to Y_2$, the map $f_*=f\circ-:\mathrm{Map}(X,Y_1)\to\mathrm{Map}(X,Y_2)$ is continuous).

$p_*$ is open. First notice that the collection of open neighborhoods $$\bigcap_{k=0}^{n-1}\;[I_{k,n},V_k]$$ form a basis of the compact open topology on $\Omega(\tilde{X})$, where $0\leq k<n$ are integers, $I_{k,n}=\left[\frac{k}{n},\frac{k+1}{n}\right]$ and the $V_k\subset \tilde{X}$ are open subset of $\tilde{X}$ that arise from evenly covered open subsets of $X$ and a local trivialization of $p$ (I can be more precise that is too vague). It is (i.e. should be) straightforward to prove (using unique path lifting and the $\epsilon$-lemma of Lebesgue) that $$p_*\left(\bigcap_{k=0}^{n-1}\;[I_{k,n},V_k]\right)=\bigcap_{k=0}^{n-1}\;[I_{k,n},p(V_k)]$$ (actually $p_*([I_{k,n},V])=[I_{k,n},p(V)]$ might already be true for $V$ as above.) Being a basis, this shows that $p_*$ is an open map, and so it is a homeomorphism. This might require $X$ to be Hausdorff and/or locally path connected.

Answer 2

Here's a nice argument for when $\Omega X$ is (or has the homotopy type of) a CW complex.

It is a general fact that $\pi_n(\Omega X)=\pi_{n+1}(X)$ and we note that $p\colon \bar{X}\to X$ induces an isomorphism on all homotopy groups except for possibly $\pi_1$ and $\pi_0$, but then that means the induced map $\Omega p\colon \Omega \bar{X}\to \Omega X$ induces isomorphisms on all homotopy groups except for possibly $\pi_0$. Note that $\bar{X}$ is simply connected so $\Omega \bar{X}$ is path connected, hence so is its image which must therefore lie in $\Omega_0 X$. So the induced map on $\pi_0$ is a bijection from a one-point set to a one-point set when the codomain is restricted to $\Omega_0 X$.

By Whitehead's theorem, as $\Omega p$ is a weak equivalence between $\Omega \bar{X}$ and $\Omega_0 X$ it must also be a homotopy equivalence.