This question is a repost from stack.exchange. It didn't get a lot of attention there. Perhaps it is badly written (or silly?). If so, I'd be happy to get comments/suggestions about that.

Let $X$ be a (nice) connected topological space. Let $LX=Map(S^1,X)$ be the free loop space and $\Omega X = Map_*(S^1,X)$ the subspace of based loops (with some choice of base point for X). Now, there is a fiber sequence $$ \Omega X \to LX\to X $$

where the map $LX\to X$ is evaluation at the base point. I am interested in the following question:

Under what general conditions this sequence splits and we have a (weak) homotopy equivalence $LX \simeq X\times \Omega X$ ?

This is of course not true in general. In the case of a classifying space $BG$ of a discrete group $G$, we have $LBG$ equivalent to a disjoint union of $BC(x)$ where $x$ runs over conjugacy classes of $G$ and $C(x)$ is its centralizer subgroup (as explained nicely in the answer to this MO question). Hence, the splitting holds if and only if $G$ is abelian.

### Partial Results

I have managed to prove this in the case where $X$ itself is of the homotopy type of a based loop space. i.e. $X\simeq \Omega Y$ for some $Y$ and a little bit more generally, when $X\simeq Map_*(W,Y)$ for some (nice) connected pointed space $W$ and some pointed space $Y$. There are several ways to see this, but the slickest proof I got is just a sequence of standard adjunctions. Denote by $X_+$ the based space obtained from $X$ by adding a disjoint base point. On the one hand,

$$ LX = Map(S^1,Map_*(W,Y))=Map_*(S^1_+,Map_*(W,Y))=Map_*(S^1_+\wedge W,Y) $$ and on the other, $$ \Omega X\times X=Map_*(S^1 \vee S^0,Map_*(W,Y))=Map_*((S^1 \vee S^0)\wedge W,Y) $$

The only difference between $S^1_+$ and $S^1\vee S^0$ is in the selection of the base point, but since $W$ is *connected*, after smashing with it we get a connected space so the choice of the base point does not matter.

Generalizing the loop space case, it is natural (I think) to ask whether this is true for a general H-space. I also have a feeling that this has something to do with vanishing of whitehead products. So I'll add a more specific version of the question:

Is it true that if all whitehead products of $X$ vanish, then the above sequence splits?