Fundamental group of the space of immersions of the 2-sphere in 3-space modulo diffeomorphisms of the first

Question

In a previous Mathoverflow question, we saw that the fundamental group of the space $Imm(S^2,\mathbb{R}^3)$ of immersions the 2-sphere in ordinary 3-space is isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}$.

We also saw that rotating the sphere 360 degrees on itself gives an element of order $2$: what I mean, and more generally, if $f_0\in Imm(S^2,\mathbb{R}^3)$ is your basepoint and if $R_t\in SO(3)$ is a path which represents the non-trivial element in $\pi_1(SO(3))\equiv \mathbb{Z}/2$, then $f_0\circ R_t$ is a path which is of order $2$ in $\pi_1(Imm(S^2,\mathbb{R}^3))$.

Of course pre-composing an immersion by a diffeo does not change the image of the immersion. So the path above is not significant in terms of the image. It is interesting, instead of considering $Imm(S^2,\mathbb{R}^3)$, to look at the quotient space $Imm(S^2,\mathbb{R}^3)/Diff(S^2)$ where $Diff(S^2)$ is the space of self-diffeomorphisms of the 2-sphere. In fact we wish to retain the orientation of the sphere (to be able to label each face of the immersed spheres in a coherent way), hence I would rather look at $$Imm(S^2,\mathbb{R}^3)/Diff^+(S^2)$$ where $Diff^+(S^2)$ denotes the set of orientation preserving diffeomorphisms of $S^2$.

I think the action of $Diff^+(S^2)$ on $Imm(S^2,\mathbb{R}^3)$ is free, i.e. that there is no immersion $f$ that is invariant by a non-trivial orientation preserving diffeo of $S^2$.

Is it true that we get a fibre bundle $Imm(S^2,\mathbb{R}^3) \mapsto Imm(S^2,\mathbb{R}^3)/Diff^+(S^2)$ with fiber $Diff^+(S^2)$?

Is it true that this leads to a short exact sequence $$1\to \pi_1 Diff^+(S^2)\to \pi_1 Imm(S^2,\mathbb{R}^3) \mapsto \pi_1 (Imm(S^2,\mathbb{R}^3)/Diff^+(S^2)) \mapsto 1$$ and that the latter is equivalent to $$0\to \mathbb{Z}/2 \to \mathbb{Z}/2\times \mathbb{Z} \to \mathbb{Z} \to 0\ ?$$ Now if you insist on looking at $Imm(S^2,\mathbb{R}^3)/Diff(S^2)$, I would also be happy to know what is its fundamental group. Note that the double cover of Boy's surface is fixed by the antipodal map so the action is non-free, so we do not get a fibre bundle to begin with. However, the map $$Imm(S^2,\mathbb{R}^3)/Diff^+(S^2) \to Imm(S^2,\mathbb{R}^3)/Diff(S^2)$$ exists (it is continuous but not a cover) and we may wonder about its action at the level of the respective fundamental groups. It is interesting to note that eversions are loops in the space on the right but not on the space on the left.

Answer 1

The action of $\text{Diff}^+(S^2)$ on $\text{Imm}(S^2,\Bbb R^3)$ is free. For pick any immersion $i$ and diffeomorphism $f$; given $x \in \Bbb R^3$ $i^{-1}(x)$ is a closed discrete (because $i$ is an immersion) set and hence finite. So if $if = i$, $f$ has finite orbits, and by this previous MathOverflow question $f$ is periodic. I claim that the group action of $\Bbb Z/n$ on $S^2$ generated by $f$ is free; Cervera-Mascaro-Michor show this in Lemma 1.3 here. So it descends to a smooth map of manifolds $S^2/\langle f\rangle \to \Bbb R^3$; this is only possible of $S^2/\langle f\rangle$ is $\Bbb{RP}^2$ and $f$ is orientation-reversing. So the action of $\text{Diff}^+(S^2)$ is free.

The same Cervera-Mascaro-Michor paper, then, shows you obtain a fiber bundle $\text{Diff}^+(S^2) \to \text{Imm}(S^2,\Bbb R^3) \to \text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$.

It is a result of Smale that the inclusion $SO(3) \to \text{Diff}^+(S^2)$ is a homotopy equivalence. Because $\pi_0$ and $\pi_2$ of $SO(3)$ are trivial, you get the short exact sequence of fundamental groups you wanted from the long exact sequence of homotopy groups of a fibration.

Lastly there is only one exact sequence up to isomorphism of abelian groups $1 \to \Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2 \to G \to 1$ and $G$ must be $\Bbb Z \to 1$. (There is only one injective homomorphism $\Bbb Z/2 \to \Bbb Z \oplus \Bbb Z/2$.)

e: $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ is the quotient of $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}^+(S^2)$ by the involution given by precomposing with the antipodal map. The fixed point set is of infinite codimension. Its complement, then, by transversality arguments, has the same fundamental group; and away from the fixed point set this is a covering map. So we obtain that $\text{Imm}(S^2,\Bbb R^3)/\text{Diff}(S^2)$ has fundamental group $\Bbb Z$, and the quotient map induces multiplication by two on the fundamental group.