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For two manifolds $M^m$ and $N^n$ with $m<n$ the Smale-Hirsch theorem says that the differential map $d:\operatorname{Imm}(M,N)\to\operatorname{Mon}(TM,TN)$ is a weak homotopy equivalence, where $\operatorname{Imm}(M,N)$ is the space of immersion from $M$ to $N$ with the compact-open $C^{\infty}$ topology and $\operatorname{Mon}(TM,TN)$ is the space of vector bundle maps which are monomorphisms on their fibers from the tangent bundle $TM$ of $M$ to $TN$. How exactly does this now imply that there is a regular homotopy from $\operatorname{id}:S^2\to\mathbb{R}^3$ to $-\operatorname{id}:S^2\to\mathbb{R}^3$? The original paper by Smale mentions the use of the Stiefel manifold $V_{3,2}$ through $\pi_2(V_{3,2})$ but I can't figure out how to get this into play with the modern formulation of the Smale-Hirsch theorem. Any help is very much appreciated.

Anyone mildly curious can also check out this movie of such an explicit eversion:

EDIT: Or could someone simply explain to me why $\operatorname{Mon}(TS^2,T\mathbb{R}^3)$ is path-connected? This must hold true since Smale claims that all immersions $S^2\to\mathbb{R}^3$ are regularly homotopic.

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up vote 2 down vote accepted

The space of maps $\operatorname{Mon}(TS^2,T\mathbb{R}^3)$ projects to the space of maps $\operatorname{Map}(S^2,\mathbb{R}^3)$ by forgetting the fiber maps (i.e. restricting to the zero sections). The fiber over a map $u$ is then the monomorphisms from $TS^2$ to $u^* T \mathbb{R}^3$. This is a locally trivial fibration with contractible base (by contracting to a constant map at $0$), hence a product. So it is homotopy equivalent to the fiber. To compute the fiber, take $u={\rm const}$. Get maps from $T S^2$ to trivial $\mathbb{R}^3$ bundle over $S^2$, which is the same as maps $T S^2$ to $\mathbb{R}^3$ monomorphic on the fibers.

This is the space of sections of a locally trivial fibration over $S^2$ with fiber the non-compact Stiefel manifold of all 2-frames in $\mathbb{R}^3$ (pick a local trivialization on $T S^2$ and see that on that trivializing chart you just need a 2-frame at over every point). This is associated fibration to $T S^2$, so is given by clutching function and since the clutching is twice the generator of $\pi_1 (V_{3,2})$, and so is trivial, the fibration is trivial. This means the space of sections is just maps $S^2$ to the fiber, which by Gram-Schmidt retracts to $V_{2,3}$. Hence $\pi_0$ is $\pi_2(V_{2,3})=0$, as wanted.

Maybe another way to replace/rephrase that last paragraph, is to note that $T S^2$ sits in the trival $\mathbb{R}^3$ bundle over $S^2$, as in $T S^2 \oplus \nu = {\rm trivial}$, and given orientations of everything (which we are implicitly assuming, actually), a monomorphism extends uniquely to a map of the $\mathbb{R}^3$-fiber to $\mathbb{R}^3$, where the vector in $\nu$ is the unit normal to the image of the original monomorphism. The space of such maps is a subset of 3 by 3 matrices, and a partial Gram-Schmidt says that the space of all invertible matrices retracts to it. The same argument in the $S^2$ family says the fiber is homotopy equivalent to the space of maps from $S^2$ to $GL(3)$, which is again by Gram-Schmidt homotopy equivalent to maps from $S^2$ to $SO(3)$. This gives same conclusion...

Also, the eversion is done in Eliashberg-Mashchev book on h-principle, in section 4.2.

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What a fantastic answer, thank you very much! – Martin Worsek Jul 5 '11 at 13:47

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