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Offhand I can think of two ways in classical homotopy theory:

  1. Show that $\pi_k(S^n)=0$ for $k\lt n$ by deforming a map $S^k\to S^n$ to be non-surjective, then contracting it away from a point not in its image. Now use the Hurewicz theorem to show $\pi_n(S^n) = H_n(S^n) = \mathbb{Z}$, which is easy to calculate with cellular homology.

  2. Use the Freudenthal suspension theorem to induct up from $\pi_1(S^1)=\mathbb{Z}$, which you can prove using (say) the universal covering space $\mathbb{R}\to S^1$.

What other ways are there to prove $\pi_n(S^n)=\mathbb{Z}$?

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this should be community wiki – Koushik Feb 8 '13 at 0:43
Am I allowed to say "countably many"? ;-) – Andrej Bauer Feb 8 '13 at 11:31
@Tom: Is there a way to generalize that argument to $n>2$? – Mike Shulman Feb 8 '13 at 16:31
The deformation argument in (1) is superfluous. Just calculate $\pi_1$ and $H_\ast$ from the cell decomposition. Then for simply connected spaces, Hurewicz shows that the minimum $i>0$ for which $H_i\ne 0$ is the same as the minimum $i>0$ for which $\pi_i\ne 0$, and that the groups coincide for this $i$. – John Pardon Feb 9 '13 at 21:53
@unknown: the deformation argument is by no means ''superfluous'', but quite crucial. You just repackage it differently (in the cellular approximation theorem). – Johannes Ebert Jun 19 '13 at 17:56

$\pi_n(S^n)=[S^n,S^n]=\lbrace$cobordism classes of framed 0-submanifolds$\rbrace$ by the Pontrjagin-Thom construction. These are collections of points (with sign) which add up to give the degree of the maps, so this set is precisely $\mathbb{Z}$.

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A modified version of your 1: If you have the Hurewicz theorem for $\pi_n(S^n)$, then you also have it for $k < n$, so you don't need the geometric argument for $\pi_k(S^n)=0$ when $k < n$.

Alternatively, after arguing geometrically that $\pi_k(S^n)=0$ when $k < n$, you can use much the same idea to show that any map $S^n\to S^n$ is homotopic to one that takes everything to the basepoint except some little disks that are mapped in a standard way. From there you can go to seeing that $\pi_n(S^n)$ is generated by the class of the identity map. And then you don't need the full strength of Hurewicz, but just the fact that all the multiples of the identity have different effects on $H_n(S^n)$. (The first part of this is pretty close in spirit to the framed cobordism argument indicated by Chris Gerig.)

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Re: your first paragraph: ah, yes, I should have known that. – Mike Shulman Feb 8 '13 at 16:31

In the paper

R. Brown and P.J. Higgins ``Colimit theorems for relative homotopy groups'', J. Pure Appl. Algebra 22 (1981) 11-41.

we generalised to all dimensions the formulation and proof due to Crowell (1959) of the Seifert-van Kampen Theorem for the fundamental group. Our generalised proof did not involve singular homology, nor simplicial approximation. Consequences were:

  1. The Brouwer Degree Theorem (determination for $ n \geqslant 1$ of $\pi_r(S^n)$ for $r \leqslant n$; and also the determination of $\pi_n(S^n \vee W)$ where $W$ is a wedge of circles (without using covering spaces);

  2. the Relative Hurewicz Theorem (as a determination of $\pi_n(X \cup CA,x)$ if $(X,A)$ is $(n-1)$-connected);

  3. J.H.C. Whitehead's determination of $\pi_2(X \cup e^2_\lambda,X,x)$ as the free crossed $\pi_1(X,x)$-module on the $2$-cells $e^2_\lambda$ (Whitehead's proof involved transversality and knot theory, and there are maybe three other proofs);

  4. a generalisation of 3. to determine $\pi_2(X \cup_f CA, X,x)$ as a crossed module induced by $f_*: \pi_1(A,a) \to \pi_1(X,x)$ (for $A,X$ connected --at present no other proof is available). Note that Whitehead's result in 3. is the case $A$ is a wedge of circles.

The full story is also given in the EMS Tract Vol 15.

The proofs very much involve the use of filtered spaces, the algebra of the multiple compositions of cubes, and the relation with the classical invariants of relative homotopy groups.

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This can be proved using the Brouwer degree. Smooth maps $f,g\colon\thinspace S^n\to S^n$ are homotopic if and only if they have the same Brouwer degree, and any continuous map is homotopic to a smooth one.

On second thoughts, this might be identical to Chris Gerig's answer. Milnor's lovely book "Topology from the Differentiable Viewpoint" contains the details.

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I think this is more or less Hopf's original argument, isn't it? – Ryan Budney Feb 9 '13 at 0:23

I think this can be made into a proof without being circular. In "Configuration spaces and iterated loop spaces," Segal proves that the group completion of the monoid of configurations of distinct unordered points in $\mathbb{R}^n$ is $\Omega^n S^n$. $\pi_0$ of this monoid is the natural numbers so its Grothendieck group is the integers. This gives $\pi_k(S^k)=\mathbb{Z}$. Since the abelianization of the braid groups is the integers, we have that $\pi_3(S^2)=\mathbb{Z}$. Since the abelianization of the symmetric groups is $\mathbb{Z}/2$, we have that $\pi_{k+1}(S^k)=\mathbb{Z}/2$ for k>2.

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This is almost circular, as the ''group completion'' theorem depends essentially on the Hurewicz theorem, to pass from a homology equivalence to a weak homotopy equivalence. – Johannes Ebert Jun 19 '13 at 18:00

A rather roundabout method for computing the fundamental group of $S^n$ comes from using Kan's loop group construction as briefly described in this answer by John Klein. The basic theory of the Kan loop group construction justifying this method can be found in Kan's article "A combinatorial definition of homotopy groups". The actual calculation of $\pi_n(\Delta^n/\partial\Delta^n)$ as the homology of the "normalized chain complex" (the complex $N_\ast$ in the linked answer) of the Kan loop group is quite easy and quick.

However, one needs to first justify --- as done in Kan's article --- that the Kan loop group of $X$ gives the homotopy of $X$ in the manner described by John Klein's answer. This requires showing that the Kan loop group is weakly equivalent to the loop space of (a fibrant replacement of) $X$. The only proofs I have seen of this fact use the Whitehead theorem that a simply connected space (or simplicial set) which has trivial reduced homology is weakly contractible.

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A stable version of Jeremy Miller's answer uses instead the Barratt-Priddy-Quillen theorem about $\Omega^\infty\Sigma^\infty S^0$ (for example, as stated in Graeme Segal's "Categories and cohomology theories"). We immediately conclude that $\pi_0(\Omega^\infty\Sigma^\infty S^0)\cong {\mathbb Z}$, as it is the Grothendieck group of $\pi_0(\coprod_{n\in{\mathbb N}} B\Sigma_n)={\mathbb N}$. By the way, we can also compute $\pi_1\cong{\mathbb Z}/2$ as the abelianization of $\Sigma_n$ for large $n$, and $\pi_2$ as the second homology group of the alternating group $A_n$ for large $n$. Applying the Freudenthal suspension theorem to $\pi_0(\Omega^\infty\Sigma^\infty S^0)\cong {\mathbb Z}$, we obtain an isomorphism between $\pi_n(S^n)$ and $\mathbb Z$ for $n\geq 2$.

This argument is weaker and more indirect than Jeremy's answer, since it only obtains stable information, and it thus requires using the Freudenthal suspension theorem.

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This is the first theorem proven in Whitehead's book ''Elements of homotopy theory'', section I.3. The proof is by a simplicial argument, which is similar to what I believe is Hopf's original argument (if the degree is $0$, take preimages of regular values of a map and deform to reduce the number of preimages).

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