From the immersion theorem to sphere inversion I am trying to understand how Smale's result that $\pi_0 imm(S^2,\mathbb{R}^3) = 0$ follows from the "main result of immersion theory".
A formal immersion between two smooth manifolds is a continuous map $f : M \to N$ together with a map of vector bundles $\psi : TM \to f^* TN$ that is fiberwise injective. The set of all formal immersions is denoted $fimm(N,M)$.  Under some dimension compactness restrictions and assuming there is no boundary, immersion theory tells us that the inclusion $imm(N,M) \to fimm(N,M)$ is a homotopy equivalence.  I am interested in the case where $N = S^2$ and $M = \mathbb{R}^3$ (in this case the above map is a homotopy equivalence).
So I want to understand why $\pi_0 (fimm(S^2, \mathbb{R}^3) = 0$.  I have a few holes in my understanding of the derivation of this that I read.  
(1) Why is the map $fimm(S^2, \mathbb{R}^3) \to map(S^2, \mathbb{R}^3)$ given by just forgetting the second map $\psi$ a fibration?
Since (1) is true, and $map(S^2, \mathbb{R}^3)$ is contractible, $fimm(S^2, \mathbb{R}^3)$ is homotopy equivalent to the fiber which is the set of vector bundle maps from $TS^2$ to the trivial bundle $S^2 \times \mathbb{R}^3$ that are injective on the fibers. Call this set $V$.   
(2) Why is $V$ homotopy equivalent to the set of orientation preserving vector bundle isomorphisms $S^2 \times \mathbb{R}^3 \to S^2 \times \mathbb{R}^3$?
From here Smale's result follows since $\pi_2(SO(3)) = 0$.  Could someone lend me a helping hand with understanding (1) and (2)? 
 A: To answer (1), the map you described is the projection map of a locally trivial fiber bundle, which is thus a fibration. To see this, consider a typical basis element of $map(S^2,\mathbb R^3)$. The description of each basis element specifies a particular coordinate chart on $S^2$ parameterized by an open set $U \subset \mathbb R^2$. The fibration restricted to that basis element is a product, whose fiber over each point is identified with the set of linear injections $\mathbb R^2 \hookrightarrow \mathbb R^3$.
To answer (2), the first thing to notice is that the embedding $S^2 \hookrightarrow \mathbb R^3$ induces a vector bundle embedding $i : TS^2 \hookrightarrow S^2 \times \mathbb R^3$. Then, given an element of $V$, let's call it $h : TS^2 \hookrightarrow S^2 \times \mathbb R^3$, notice that there exists a unique factorization of $h$ of the form
$$TS^2 \xrightarrow{i} S^2 \times \mathbb R^3 \xrightarrow{H} S^2 \times \mathbb R^3
$$
such that $H$ is an isomorphism on each fiber. The map $H$ is defined on $\{x\} \times \mathbb R^3$ to be the unique orientation preserving linear isomorphism which extends $h_x$ and takes the normal line of the plane $T_x S^2 \subset \mathbb R^3$ isometrically to the normal line of the plane $h(T_x(S^2)) \subset \mathbb R^3$. So $V$ therefore embeds naturally into the set of vector bundle isomorphisms $S^2 \times \mathbb R^3 \mapsto S^2 \times \mathbb R^3$.
So now you just have to check that the image of this embedding is a deformation retraction, which you can do by applying a kind of Gram-Schmidt argument to the normal directions of $h(T_x S^2)$ for each $x \in S^2$.
