Homotopy of space of immersions, Smale-Hirsch theorem

Question

If $M$ and $N$ are simply connected manifolds with $\dim M< \dim N$, we denote by $Imm\left(M,N\right)$ the space of immersions of $M$ in $N$.

Let $M$ and $M'$ manifolds of dimensions $m>0$. It is true that if $M$ is homotopic to $M'$, then for $k\geq m$, the spaces $Imm\left(M,\mathbb{R}^{m+k}\right)$ and $Imm\left(M',\mathbb{R}^{m+k}\right)$ are homotopic? i.e

if $k\geq m$, then $M\simeq M'\Rightarrow Imm\left(M,\mathbb{R}^{m+k}\right)\simeq Imm\left(M',\mathbb{R}^{m+k}\right)$?

Thanks

Abdoul

Answer 1

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Tom Goodwillie below.

No. For example, if $M$ is a Moebius band then, at least for even $k$, $Imm(M,\mathbb R^{2+k})$ is not homotopy equivalent to $Imm(S^1\times \mathbb R,\mathbb R^{2+k})$.

The latter is equivalent to the space of all maps from $S^1$ to $O(2+k)/O(k)$. Its first non-trivial homotopy group is $\pi_{k-1}=\pi_k(O(2+k)/O(k))=\mathbb Z$.

The former is equivalent to the space of sections of a nontrivial bundle with base $S^1$ and fiber $O(2+k)/O(k)$, in which the action of the fundamental group of the base on $\pi_k(O(2+k)/O(k))$ is nontrivial. Its first non-trivial homotopy group is $\mathbb Z/2\mathbb Z$.