Possible mistake in Cohen notes "Immersions of manifolds and homotopy theory" (version 27 March 2022)

Question

In Theorem 2 of these notes, Ralph Cohen reformulates the main theorem of Hirsch-Smale theory merely in terms of normal bundles. In particular, he says that if $N, M$ are two manifolds, $\dim N< \dim M$ then two immersions $f_1, f_2:N\to M$ are regularly homotopic if and only if their normal bundles $\nu(f_1)$ and $\nu(f_2)$ are isomorphic.

Why it is enough to have isomorphic normal bundle to have regularly homotopic immersions? In other words, how does the if part of the statement follows from the classical formulation of the Hirsch-Smale theorem?

I recall that the celebrated Hirsch-Smale theorem states that if $\dim N< \dim M$ or $N$ open, then the tangential map gives a bijection between the connected components $$\pi_0(Imm(N,M)) \to \pi_0(Mono(TN,TM))$$ where $Imm(N,M) $ is the set of immersions and $Mono(TN,TM)$ is the set of monomorphisms (bundle maps $TN\to TM$ injective fiberwise).

My thoughts

Now, assume that $\nu(f_1)\simeq \nu(f_2)$ are isomorphic to some $\nu\to N$. In view of Hirsch-Smale's theorem, we would like to show that $df_1: TN\to TM$ and $df_2:TN\to TM$ are homotopic through monomorphisms. The hypothesis imply the existence of two isomorphisms $$F_i: TN\oplus \nu \to TM$$ for $i=1,2$, where $F_i|_{TN} = df_i$. Clearly, if we show that $F_1$ and $F_2$ are homotopic through isomorphisms, then $df_1$ and $df_2$ are homotopic through monomorphisms so we are done. However I don't see any good reason why $F_1$ should be homotopic to $F_2$. Notice that we only know that $\nu(f_1)\simeq \nu(f_2)$ but we have some freedom in choosing the embedding $\nu\to TM$, constructing different $F_1, F_2$. It is enough to show that any of these are homotopic.

A special case

Let's study a special case, when $N =\mathbb S^n$ and $TM\simeq \varepsilon^m$ is trivial. Then we are given two trivializations $F_1, F_2$ of $TN\oplus \nu$, and we would like to show that these are homotopic (through trivialization). An (oriented) trivialization is a section of the bundle of oriented $n+v$-frames, $SO(TN\oplus \nu)\to N$, where $v= \mathrm{rank} (\nu)$. Which has fiber $SO(n+v)$. The obstructions to construct an homotopy lie $H^i(\mathbb S^n, \pi_i(SO(n+v))$, thus the obstruction to homotoping $F_1$ to $F_2$ lies in $\pi_n((SO(n+v)))$ (provided the two are cooriented). However the latter homotopy group is in general non-trivial. As I said above, we have some freedom, i.e. we can change the trivialization $F_1:TN\oplus \nu \to \varepsilon^m$, by twisting it with an automorphism of $Aut(\nu)$, thus to conclude it would be sufficient to show that $\pi_n(SO(v))$ acts transitively on $\pi_n(SO(n+v))$ where the action is induced by multiplication by $A\in SO(v)\subset SO(n+v)$ (the inclusion is $A\to \begin{bmatrix} 1 & 0\\ 0& A\end{bmatrix}$).

Answer 1

The statement is false in two ways. First, two immersions might not even be homotopic even though their normal bundles are both, say, trivial. Second, even if $M=\mathbb R^m$, regular homotopy classes of immersions of $N$ correspond to homotopy classes of vector bundle monomorphisms from $TN$ to the trivial rank $m$ bundle. Two such monomorphisms can lead to isomorphic normal bundles without being homotopic. You can already see counterexamples with $n=S^1$ and $m=2$.