5
$\begingroup$

I'am trying to come up with a solution to the referred problem which, by the way, states the following:

If $m<p$, show that every map $M^m\to S^p$ is homotopic to a constant map.

From the chapter, you may (must) assume $M$ is a compact (and, may be, boundaryless) smooth ($=C^\infty$) manifold (Hausdorff, second countable, locally euclidean topological manifold of dimension $m$). I've been trying to use Milnor's Theorem B:

Two mappings from $M$ to $S^p$ are smoothly homotopic if and only if the associated Pontryagin manifolds are framed cobordant.

I can't come up with a framed manifold corresponding to a constant function. Does it make any sense? I mean, what are the the regular values of such functions? What can you do with an empty set, in this case?

I've lots of questions here, if all I said above is completely nonsense, how to proceed then? In what other books could I learn more (I've googled a lot and all other references seemed to copy Milnor's work...)

Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ Do you know Sard's theorem? $\endgroup$ – Tim kinsella Aug 8 '16 at 1:52
  • $\begingroup$ It asserts that the set $f(M)$ has measure zero in $S^p$, right? Still don't get it, but already saw I can't apply Theorem B I was speaking above due to the inequality $m<p$. $\endgroup$ – user255306 Aug 8 '16 at 1:58
  • $\begingroup$ Yes, thats how you show $f$ is not surjective in the first line of Tsemo's answer below. $\endgroup$ – Tim kinsella Aug 8 '16 at 2:14
4
$\begingroup$

If $m<p$, then the image of $f$ is not surjective, let $y\in S^p$ which is not in $f(M)$ and $p_y:S^p\rightarrow R^p$ the stereographic projection centred at $y$, $p_y\circ f$ is homotopic to a constant: define $H_t(x)=t(p_y\circ f)(x)$, $p_y^{-1}\circ H_t$ is the requested homotopy.

I use the fact that the stereographic projection $p_y:S^p-\{y\}\rightarrow R^p$ is a diffeomorphism

$\endgroup$
  • $\begingroup$ All right, man! This puts an end to these question. Thanks, everybody, for the help! $\endgroup$ – user255306 Aug 8 '16 at 3:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.