A fundamental result in three-dimensional smooth topology, which in computer jargon we might refer to as "a primitive", is the statement that any ($C^\infty$) diffeomorphism of the two-sphere $S^2$ extends to a diffeomorphism of the closed three-ball $D^3$. Equivalently: $\mathrm{Diff}(S^2)$ is connected. This theorem was first proven by Munkres [Mich. Math. Jour. 7 (1960), 193-197]. Later, Smale proved the stronger result that $\mathrm{Diff}(S^2)$ has the homotopy type of $O(3)$ [Proc. AMS 10 (1959), 621-626]. Another proof of Smale's result is given by Cerf in the appendix to [Sur les difféomorphismes de la sphère de dimension trois ($Γ\_4=0$), Lecture Notes in Mathematics, No. 53. Springer-Verlag, Berlin-New York 1968].  




> 
> **Question 1:** Are there any other known proofs of the statement that any diffeomorphism of the two-sphere $S^2$ extends to a diffeomorphism of the closed three-ball $D^3$?
> 


There are two reasons I'm not fully happy with the proofs I cited above. Smale's proof and Cerf's proof show much more and use what looks to me like "too much machinery" for just the "$\mathrm{Diff}(S^2)$ is connected" statement, and, in particular, machinery which seems outside basic differential topology (maybe I'm wrong; I haven't gone into them in much detail). Munkres's proof has a number of back-references to another of his papers [Ann. Math. 72(3) (1960), 521-554], and corners need to be smoothed over and over and over and over again to get an honest smooth isotopy between a given diffeomorphism of $S^2$ and the identity. What is worse, it seems difficult to extract an algorithm from Munkres's proof (Lemma 1.1 looks non-constructive - I wouldn't know how to extract a concrete diffeomorphism out of its proof), which brings me to my second question:



> 
> **Question 2:** How could I **implement** an extension of a smooth diffeomorphism of the two-sphere to the three-ball? To make things really concrete, let's say I had an image of the surface of the earth which I deformed by some strange diffeomorphism $f$ of $S^2$. How (by computer) could I smoothly deform it back to the usual picture of the earth? 
> 


One dimension down, maybe one way to do it might be to "relax a diffeomorphism of a circle gradually using the heat equation" (see Greg Kuperberg's comment [here](http://gowers.wordpress.com/2011/03/23/milnor-wins-2011-abel-prize/)). Does this work one dimension up? I couldn't figure this out, but I don't see an obvious obstruction- not in dimension three. Or maybe there's a slick way of implementing Munkres's proof by lifting an orientation-preserving diffeomorphism of $S^2$ to $\mathrm{Spin}(3)$ or something... I really have no idea.  



Note, though, that other proofs that diffeomorphisms of $S^1$ extend to $D^2$ clearly seem to fail in dimension three... in particular, trying to use some sort of Alexander trick to comb all the "bad parts" of the diffeomorphism into a small disc and shrink that disc to a point will not give rise to a smooth isotopy.  



Finally, Morris Hirsch says in a footnote on Page 38 of The Collected Papers of Stephen Smale: "Around this time [1959] an outline of a proof attributed to Kneser was circulating by word of mouth; it was based on an alleged version of the Riemann Mapping Theorem which gives smoothness at the boundary of smooth Jordan domains, and smooth dependence on parameters. I do not know if such a proof was ever published."



> 
> **Question 3:** Was such a proof ever published? Is there anything else to be said about this proof outline?
> 


**Edit:** Actually, I'd like to add even a fourth question:



> 
> **Question 4:** Are there any "second generation" detailed expositions of any of the above proofs?
> 



Extending a diffeomorphism of the sphere $S^2$ to the ball $D^3$
More a survey of related things than an answer, but here goes.


Let's write $D(n)$ for the space of compactly supported diffeomorphisms $\mathbb R^n\to \mathbb R^n$. A reasonable guess might be that this is contractible, but this is true only for very small values of $n$.


The space $Diff(S^n)$ of diffeomorphisms $S^n\to S^n$ contains the Lie group $O(n+1)$. A reasonable guess might be that the inclusion of $O(n+1)$ is a homotopy equivalence, but no. In fact, this is equivalent to the first guess: The subgroup of $Diff(S^n)$ consisting of diffeomorphisms supported in the complement of a given point may be identified with $D(n)$, and the multiplication map $D(n)\times O(n+1)\to Diff(S^n)$ (which is not a group homomorphism) is an equivalence. You can see this by comparing $O(n+1)$ with the space of cosets $Diff(S^n)/D(n)$. Thus $D(n)$ is what you might call the exotic part of the homotopy type of $Diff(S^n)$. It is also equivalent to the space of all diffeomorphisms $D^n\to D^n$ fixing the boundary pointwise.


Introduce one more player: the space of all compactly supported diffeomorphisms $\mathbb R^n\times \lbrack 0,\infty )\to \mathbb R^n\times \lbrack 0,\infty )$. Call this $P(n)$. It fibers over $D(n)$, and the fiber is equivalent to $D(n+1)$. It is equivalent to the space of "pseudoisotopies" of $D^n$.


The statement that every diffeomorphism of $S^n$ extends to $D^{n+1}$ means precisely that $P(n)\to D(n)$ induces a surjection on components. This is in principle weaker than the statement that $D(n)$ is connected, but it turns out that (at least for most values of $n$, maybe all?) $P(n)$ is connected.


In low dimensions the story is this:


$D(0)$ is a point.


$P(0)=D(1)$ is contractible because it is convex: convex linear combinations of order-preserving 
diffeomorphisms of the line are again order-preserving diffeomorphisms.


$P(1)\sim D(2)$ is contractible. I am aware of two approaches to this:


(1) I believe that when Smale (re-)proved this he used the Poincare-Bendixson Theorem. The crux is that, given a compactly supported field of tangent lines in the half-plane $y\ge 0$ transverse to the line $y=0$, if you follow it from $y=0$ you will get all the way up and not get trapped in some spiral. 


(2) Complex analysis. I always imagine that the following works, but I'm not sure of the details: The space of Riemannian metrics on $S^2$ is contractible because it is convex. The space of conformal structures on $S^2$ is contractible because its product with the space of positive functions on $S^2$ is that space of metrics. The group $Diff(S^2)$ acts on this contractible space. It acts transitively, by the uniformization theorem, and presumably in a strong enough sense that this implies that the subgroup preserving the standard conformal structure is equivalent to the whole. This subgroup (Moebius transformations plus complex conjugation) is equivalent to its maximal compact subgroup $O(3)$.


Thus $P(2)\sim D(3)$. Smale conjectured that this is contractible. Hatcher proved it.


Thus $P(3)\sim D(4)$. I don't actually know anything about this space.


For large values of $n$ (I forget how large):


The space $P(n)$ is connected, by a theorem of Cerf. But this does not mean at all that $D(n)$ is connected: we have an exact sequence $\dots \to\pi\_1D(n)\to \pi\_0D(n+1)\to \pi\_0P(n)\to \pi\_0D(n)$.


$\pi\_0D(n)$ is the Kervaire-Milnor group of homotopy $(n+1)$-spheres, known to be finite and frequently nontrivial: You can make a smooth manifold homeomorphic to $S^{n+1}$ from any element of $\pi\_0D(n)$ by gluing two hemispheres together. By the $h$-cobordism theorem, you get all homotopy spheres in this way; and it's elementary to see that two elements give the same thing if and only if they differ by something in the image of $\pi\_0P(n)$.


There is an important map $P(n)\to P(n+1)$. Hatcher showed, and Igusa proved, that it is about $\frac{n}{3}$-connected. In the stable range $P(n)$ is essentially the Waldhausen $K$-theory of a point, which is rationally the same as the algebraic $K$-theory of $\mathbb Z$, and this implies plenty of elements of infinite order in the homotopy groups of $P(n)$ and $D(n)$ in degrees less than about $\frac{n}{3}$.


