Paths and Homotopy

Sphere Eversion: I · II · III · IV · V · VI

Homotopy is continuous deformation. This volume builds it from the path of Vol. I and computes the two facts about spheres that Smale’s proof will need.

Vol. I restated the problem: an eversion is a path in the space $\operatorname{Imm}(S^2,\mathbb{R}^3)$. A path in a space of maps is a continuously varying family of maps. That has a name.

Homotopy

The notation $X\times[0,1]$ matters. For $X=S^2$ it is a thickened sphere, a spherical shell: each point of the sphere grows a little time-interval, pointing in a new direction.

Any two maps $X\to\mathbb{R}^n$ are homotopic, by the straight-line homotopy

The same formula works in any convex set, because the segment between $f(x)$ and $g(x)$ stays inside. Homotopy only becomes interesting when the target has a hole for the segment to fall into. Below, $\gamma_{0}$ and $\gamma_{1}$ are two paths from $A$ to $B$ and the orange curve is the frame $H_{s}$.

$H(t,s)=(1-s)\,\gamma_{0}(t)+s\,\gamma_{1}(t)$, with both endpoints held fixed. In the plane this always works. With the origin removed, for these two paths, it works exactly when the loop "$\gamma_{0}$ then $\gamma_{1}$ backwards" has winding number $0$ around the hole. Drag the green handle across the hole and watch both facts change together.

When the endpoints are held fixed throughout, as here, the homotopy is called a homotopy rel endpoints. That is the right notion for paths; without it, any path could be reeled in to a constant one.

Loops and the fundamental group

A loop at $x_{0}$ is a path that starts and ends at $x_{0}$. Two loops can be concatenated: run the first at double speed, then the second.

(A group is a set with an associative product, an identity element, and an inverse for every element.) Checking the group laws is a matter of reparametrizing time, and every reparametrization is itself a homotopy. A space in which every loop is homotopic to a constant (and which is path-connected) is simply connected: $\pi_{1}=0$. A set is convex if it contains the segment between any two of its points, and every convex set is simply connected, by the straight-line homotopy. More generally a space is contractible if its identity map is homotopic to a constant map: the whole space can be shrunk to a point inside itself. $\mathbb{R}^n$ is contractible, and a contractible space has no interesting loops, or spheres of any dimension.

The widget above shows the first nontrivial example. In $\mathbb{R}^2\setminus\lbrace 0\rbrace$, the loop “$\gamma_{0}$ then $\gamma_{1}$ backwards” is homotopic to a constant exactly when its winding number is $0$. To prove that, we need to define winding number properly, and the cleanest place to do that is the circle.

The circle: π₁(S¹) = ℤ

The map

wraps the line around the circle infinitely often. It is a covering map: every small arc $U$ of the circle has a preimage that is a disjoint union of copies of $U$ (one per integer), each mapped homeomorphically onto $U$. Picture the line as a helix sitting over the circle, with $p$ as vertical projection.

Path lifting. For every path $\gamma$ in $S^1$ starting at $p(0)$ there is exactly one path $\tilde\gamma$ in $\mathbb{R}$ with $\tilde\gamma(0)=0$ and $p\circ\tilde\gamma=\gamma$.

Why. By compactness of $[0,1]$ (Vol. I), chop time into finitely many pieces, each of which $\gamma$ maps into one small arc. Over a small arc, the covering is a stack of copies, and you have no choice but to stay on the copy you started in. Glue the pieces. $\square$

Homotopy lifting. The same argument lifts homotopies. Chop the square $[0,1]\times[0,1]$ into a grid fine enough that each small square lands in one small arc (compactness again), then lift the small squares one at a time, row by row. Each new square meets what is already lifted along its left and bottom edges, a connected set, and that pins down which copy of the arc it must use.

Now define the degree of a loop $\gamma$ at $p(0)$ to be $\tilde\gamma(1)$. It is an integer, because $p(\tilde\gamma(1))=\gamma(1)=p(0)$. A homotopy of loops lifts to a homotopy of lifts, along which the endpoint moves continuously in $p^{-1}(p(0))=\mathbb{Z}$. By Vol. I’s corollary, it cannot move. So:

(An isomorphism of groups, written $\cong$, is a bijection that respects the products; isomorphic groups are the same group with the elements renamed. Here $\deg$ respects products because the lift of $\gamma$ followed by $\delta$ is $\tilde\gamma$ followed by $\tilde\delta+\deg\gamma$. It is onto because $t\mapsto p(nt)$ has degree $n$, and one-to-one because two loops with the same lift endpoint have lifts related by a straight-line homotopy in the convex set $\mathbb{R}$, which projects down.)

The helix is $\mathbb{R}$, coiled so that $p(s)=(\cos 2\pi s,\sin 2\pi s)$ is "drop straight down onto the circle". The blue point runs a loop $\theta(t)=2\pi\bigl(nt+w\sin 2\pi t\bigr)$ on the circle; the orange point is its unique lift starting at $0$. However much you wiggle, the lift ends exactly on the integer $n$. That integer is the class of the loop in $\pi_{1}(S^1)$. Drag to turn the picture.

The winding number of a loop $\gamma$ in $\mathbb{R}^2\setminus\lbrace 0\rbrace$ is the degree of $\gamma/\lVert\gamma\rVert$, a loop on the unit circle. The straight-line homotopy $(1-s)\gamma+s\,\gamma/\lVert\gamma\rVert$ never hits $0$, so $\mathbb{R}^2\setminus\lbrace 0\rbrace$ and $S^1$ have the same loops up to homotopy: $\pi_{1}(\mathbb{R}^2\setminus\lbrace 0\rbrace)\cong\mathbb{Z}$. That is the missing proof for the first widget.

Keep this pattern in mind; it is the core of Vol. V in miniature. A covering map $E\to B$ with discrete fibre $F$, the preimage of a point (here $\mathbb{R}\to S^1$, fibre $\mathbb{Z}$), plus the fact that $E$ has no interesting loops, computes the loops of $B$ from the points of $F$.

The sphere: π₁(S²) = 0

Now go up a dimension. Take a loop on $S^2$. If it misses some point $q$, then it lives in $S^2\setminus\lbrace q\rbrace$, and stereographic projection from $q$ is a homeomorphism $S^2\setminus\lbrace q\rbrace\to\mathbb{R}^2$. From the north pole it sends $(x,y,z)$ to $\bigl(x/(1-z),\,y/(1-z)\bigr)$, the point where the line from the pole through $(x,y,z)$ meets the plane $z=0$. In $\mathbb{R}^2$ the straight-line homotopy shrinks the loop to a point. Project back.

A loop drawn around the removed south pole. On $S^2\setminus\lbrace S\rbrace$ it slides up over the equator and shrinks to a point at the top: the sphere minus a point is a plane in disguise, and every loop in it is contractible. Remove the north pole too and the space becomes a cylinder; the same loop is caught between the two holes.

There is a real gap in that argument. A continuous loop can hit every point of $S^2$: space-filling curves exist. The repair is to replace the loop by a tame one. By uniform continuity (Vol. I), chop time into $N$ pieces so short that $\gamma$ moves less than $\tfrac12$ within each. On each piece, let $s(t)$ run along the straight segment in $\mathbb{R}^3$ between the endpoints’ images, at the same rate. Then $s(t)$ is within $\tfrac12$ of $\gamma(t)$, so the whole segment from $\gamma(t)$ to $s(t)$ stays at least $\tfrac12$ away from the origin, and pushing it out onto the sphere by $x\mapsto x/\lVert x\rVert$ gives a homotopy on $S^2$ from $\gamma$ to the loop $\beta=s/\lVert s\rVert$. (At $t=0$ nothing moves, so the base point stays put.) Now $\beta$ is $N$ short arcs, and an arc covers no area: cut it into $m$ equal pieces, each fits in a cap of radius $c/m$ for some constant $c$, and the $m$ caps have total area at most about $m\cdot\pi(c/m)^2=\pi c^2/m$, which goes to $0$ as $m$ grows. Finitely many arcs cannot cover the $4\pi$ of area on the sphere, so $\beta$ misses a point, and stereographic projection finishes. (This is the easy case of Sard’s theorem: a smooth image of something lower-dimensional has zero volume.) Therefore

Compare the circle: there the loop that goes once around cannot miss a point, which is why this argument cannot shrink it.

Higher homotopy groups

A loop is a map from $S^1$. Replace $S^1$ by $S^k$ and you get the higher homotopy groups:

The argument for $\pi_{1}(S^2)=0$ works word for word in every dimension where the domain is smaller than the target:

Proof. The repair above works in any dimension. Treat $f:S^k\to S^n$ as a map of the cube $[0,1]^k$, and chop the cube into $N^k$ small cubes on each of which $f$ moves less than $\tfrac12$. On each small cube replace $f$ by the polynomial that matches it at the corners and is linear in each coordinate separately, pushed out onto $S^n$. Its values are averages of corner values, all within $\tfrac12$ of $f$, so as before the pushed-out straight-line homotopy joins $f$ to this tame map $\beta$; on the boundary of the big cube every corner value is the base point, so $\beta$ is too. Now cut each small cube into $m^k$ pieces. Each piece lands in a ball of radius $c/m$, so the image of $\beta$ is covered by $N^km^k$ balls of $n$-dimensional volume about $(c/m)^n$ each, a total of order $m^{k-n}$, which goes to $0$ as $m$ grows because $k<n$. So $\beta$ misses a point $q$. Stereographic projection from $q$ carries $S^n\setminus\lbrace q\rbrace$ to $\mathbb{R}^n$, where the straight-line homotopy contracts $f$ (to the base point, if we contract toward it). $\square$

The instance Smale’s proof needs, in Vol. V, is

A 2-sphere inside a 3-sphere always has room to shrink. When the dimensions match, the answer is instead $\pi_{n}(S^n)\cong\mathbb{Z}$, the degree again: how many times, with sign, the map covers the target. We met the $n=1$ case above; the $n=2$ case is the degree of the Gauss map in Vol. IV. And a warning, to show the theorem above is not “obvious”: when the domain is bigger, things are wild. $\pi_{3}(S^2)\cong\mathbb{Z}$, generated by the Hopf fibration, a map from the 3-sphere onto the 2-sphere that no homotopy can undo. In the quaternions of Vol. V it is $q\mapsto qiq^{-1}$: send a unit quaternion to the point where its rotation carries $(1,0,0)$. Every point of $S^2$ has a whole circle of preimages, and any two of those circles are linked like two rings of a chain.

Where this goes

Two computations to carry forward, and one principle:

1. $\pi_{1}(S^1)=\mathbb{Z}$, via the covering $\mathbb{R}\to S^1$. It will reappear as the turning number of a plane curve in Vol. IV.
2. $\pi_{2}(S^3)=0$, via Sard and stereographic projection. It will be the last link in the proof in Vol. V.
3. Homotopy invariants are discrete, and discrete things cannot move continuously. That is Vol. I’s corollary, and every obstruction in this series is an instance of it.

But nothing so far knows about smoothness. Homotopy lets a sphere be crushed to a point, folded flat, and unfolded. To say what is forbidden in an eversion we need derivatives, and a sphere to take them on.