TITLE: Classification of sections of free loop fibration over the two-sphere
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: The space of sections of $\Lambda S^2\to S^2$ is a subspace of the space of all maps 
$$S^2\to \Lambda S^2=Map(S^1,S^2).$$
When viewed as a subspace of the space of all maps 
$$S^1\to Map(S^2,S^2),$$
it becomes $\Omega_1 Map(S^2,S^2)$, the based loopspace of the space of self-maps of $S^2$, with the identity map as basepoint. Thus the problem is to compute $\pi_1(Map(S^2,S^2),1)$. I claim that it has order two.
The space of all degree one maps $S^2\to S^2$ is sometimes denoted by $SG(3)$; it is the "structure group" for oriented spherical fibrations of rank $3$. It is part of a fibration sequence
$$
SF(2)\to SG(3)\to S^2
$$
with $SF(2)$ the space of degree one based maps $S^2\to S^2$, a component of $\Omega^2S^2$. Mapping into this is another fibration sequence 
$$
SO(2)\to SO(3)\to S^2.
$$
Thus $SG(3)/SO(3)$ is homotopy equivalent to $SF(2)/SO(2)$. 
The latter is $1$-connected and has finite $\pi_2$. This follows as soon as one knows that $\pi_1SO(2)\to \pi_1SF(2)=\pi_3S^2$ is an isomorphism, which I think is not hard to see.
Thus $\pi_1SG(3)$ is the cokernel of the map $\pi_2(SG(3)/SO(3))\to \pi_1SO(3)=\mathbb Z/2$.
This map is zero because it factors through a map $\pi_2(SG(3)/SO(3))\to \pi_1SO(2)$ from a finite group to $\mathbb Z$.