Let m, n > 1. Is it true that C(S^m, S^m), and C(S^n, S^n) are homeomorphic ? [both endowed with the sup metric (or equivalently the compact-open topology)]

Generally, C(S^n, S^n), with n >= 1, is a countably infinite (disjoint) union of path-connected (due to Hopf) components C_{n,k}, k in Z.

I think each of these components may be viewed as an infinite dimensional Frechet manifold. Unfortunately, they are not contractible. However, the question is [somewhat vaguely] motivated by the Henderson's Theorem.

Also, I have some related questions : - Fixing n, are the C_{n,k}'s homeomorphic to each other (at least for k <> 0) ? - Are there some m, n, k, l with m <> n s.t. C_{m,k} ~ C_{n,l} ?

What I was able to do in this direction until now is to show the existence of a proper, one-to-one, degree-preserving map from C(S^m, S^m) into C(S^n, S^n). Even for m >> n, but far to be surjective.

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    $\begingroup$ What is C(S^m,S^m) ? the space continuous functions or something else? $\endgroup$ – Ryan Budney Dec 13 '09 at 3:04
  • $\begingroup$ Yes. It is the space of all continuous self-maps of S^m. $\endgroup$ – Ady Dec 13 '09 at 3:10
  • $\begingroup$ Guessing wildly here, but here's my suspicion: The C_{n,n} have component retracting to $S^n$ --- the index-zero maps; think of $S^n$ as the images of the north pole --- and the complement should deformation-retract to $\coprod_\infty SL_n(\mathbb{R})$ or some such thing; and the $SL_n$s certainly aren't homeomorphic. Of course, guessing wildly isn't necessarily helpful. $\endgroup$ – some guy on the street Dec 13 '09 at 18:43
  • $\begingroup$ @Ady: when someone asks for clarification in the comments (especially in a case like this, where the clarification is necessary), it's best to both answer in the comments and edit your question to improve the exposition. $\endgroup$ – Scott Morrison Dec 13 '09 at 18:47

I see no reason for $C(S^m, S^m)$, $C(S^n, S^n)$ to be homeomorphic. Here is a proof that their identity components are not homeomorphic.

The identity component of $C(S^m, S^m)$ is usually denoted $SG_{n+1}$. The homotopy fiber $SF_n$ of the evaluation map $SG_{n+1}\to S^n$ that sends a map to the image of the basepoint is homotopy equivalent to a component in the $n$-fold loops space on $S^n$. So $\pi_k(SF_n)=\pi_{k+n}(S^n)$, and the homotopy sequence of the fibration shows that if $n$ is odd, then the evaluation map is a rational isomorphism, in particular, $\pi_k(SG_{n+1})$ is infinite if and only if $k=n$.

If memory serves me it is not hard to compute rational homotopy type of $C(S^m, S^m)$. The homotopy type is harder to understand, but look for works of Vagn Lundsgaard Hansen (search in mathscinet with sphere in the title field).


  1. Ady: $SG_{n+1}$ is not the set of pointed maps, it consists of maps homotopic to identity.

  2. I did not claim it enough to show that identity components are not homeomorphic. My answer was intended to push you in the right direction not to research the whole matter for you.

  3. Reid Barton: I think knowing that different components of the $\Omega^nS^n$ are homotopy equivalent does NOT imply that different comonents of $C(S^n, S^n)$ are homotopy equivalent.

  4. If memory serves me, components of $C(S^n, S^n)$ are NOT all homotopy equivalent as shown eg by Hansen in the papers I mentioned such as [Hansen, Vagn Lundsgaard, The homotopy problem for the components in the space of maps on the $n$-sphere. Quart. J. Math. Oxford Ser. (2) 25 (1974), 313--321].

ONE MORE REFERENCE: Lupton, Gregory; Smith, Samuel Bruce, Criteria for components of a function space to be homotopy equivalent. Math. Proc. Cambridge Philos. Soc. 145 (2008), no. 1, 95--106.

  • $\begingroup$ I can agree C(S^n, S^n) (as a space of pointed maps) has the same homotopy groups as S^n just shifted by n dimensions. But the original question was about C(S^n, S^n) as a whole, i.e., no basepoint, no pointed maps. Also, if two components are not homeo, does it follow the two spaces are not homeo ? $\endgroup$ – Ady Dec 13 '09 at 4:39
  • $\begingroup$ Why is it enough to show the identity components are not homeomorphic? $\endgroup$ – Mariano Suárez-Álvarez Dec 13 '09 at 5:21
  • $\begingroup$ Also, what about the second related question ? $\endgroup$ – Ady Dec 13 '09 at 5:32
  • $\begingroup$ By the way, I deleted my comment because after reading Igor's post more carefully, I realized that everything I said in it is basically contained in his post. $\endgroup$ – Andy Putman Dec 13 '09 at 6:04
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    $\begingroup$ $SF_n = \Omega^n S^n$ is a loop space, so its components are all homotopy equivalent, and therefore your argument should apply to any component of $C(S^n, S^n)$. $\endgroup$ – Reid Barton Dec 13 '09 at 7:21

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