For any space $X$ there is a fibration $$ \Omega X\to LX\stackrel{ev}{\to} X $$ where $LX=Map(S^1,X)$ is the free loop space, $\Omega X = Map_*(S^1,X)$ is the based loop space, and $ev:LX\to X$ is the evaluation map sending $\omega$ to $\omega(1)$. There is a canonical section $\sigma_0:X\to LX$ sending $x\in X$ to the constant loop at $x$.

I'm interested in the classification of sections of this fibration up to fiberwise homotopy, which seems to be a difficult problem in general. In particular, the standard obstruction theory methods (such as Corollary VI.6.16 on p.302 of Whitehead's "Elements of homotopy theory") don't seem to apply.

So I'm asking about what I believe to be the simplest example where this fibration does not split, namely $X=S^2$.

Have the sections of $ev: LS^2\to S^2$ been classified up to fiberwise homotopy?

The only reference I could find on this problem was

*Samson Saneblidze*, MR 1255935 **On the homotopy classification of sections in the free loop fibration**, *J. Pure Appl. Algebra* **91** (1994), no. 1-3, 317--327.

from which I can probably deduce an answer for the rationalization $S^2_\mathbb{Q}$.

In case the answer turns out to be "no", let me ask an easier question. There is an action $S^1\times S^2\to S^2$ given by rotation around an axis. The adjoint of this action gives a section $\sigma: S^2\to LS^2$.

Is the section $\sigma$ fiberwise homotopic to the canonical section $\sigma_0$? If not, how do I see this?

**Added later:** I think I see a way to distinguish between $\sigma$ and $\sigma_0$. Denote the adjoints of these sections by $\sigma',\sigma_0':S^1\times S^2\to S^2$. Let $S^1\subseteq S^2$ be a diameter which passes through the two fixed points of the rotation. Then $\sigma'$ and $\sigma_0'$ when restricted to $S^1\times S^1\subset S^1\times S^2$ have different degrees (1 and 0), hence they are not homotopic.

I wonder if this can be formalized somehow.