Sphere Eversion: I · II · III · IV · V · VI
One dimension down, eversion is impossible, and the reason is an integer. This volume finds the integer, proves it cannot change, and shows exactly what goes wrong when you cheat.
Recall from Vol. III: an immersion is a smooth map whose derivative never drops rank, and a regular homotopy is a path of immersions. Everting $S^2$ means joining $\iota(p)=p$ to $\alpha(p)=-p$ by such a path. Before trying, do the same thing for $S^1$ in $\mathbb{R}^2$, with one adjustment: there $p\mapsto-p$ is a rotation by $\pi$, so the map that turns the circle inside out is a reflection, $(x,y)\mapsto(x,-y)$.
Circling in one dimension less
An immersed closed curve $\gamma:S^1\to\mathbb{R}^2$ has a unit tangent $\mathbf{T}(t)=\gamma’(t)/\lVert\gamma’(t)\rVert\in S^1$. The turning number is how many times $\mathbf{T}$ wraps the unit circle,
\[\tau(\gamma)=\frac{1}{2\pi}\bigl(\theta(2\pi)-\theta(0)\bigr)=\frac{1}{2\pi}\int_{S^1}d\theta,\qquad \mathbf{T}=(\cos\theta,\sin\theta).\]For $\gamma(t)=(\cos t,\sin t)$ one has $\mathbf{T}(t)=(-\sin t,\cos t)$ and $\tau=+1$. For the reflected parametrization $\bar\gamma(t)=(\cos t,-\sin t)$ one has $\tau=-1$. Turning number is an integer and varies continuously under regular homotopy, so it cannot jump. Therefore $\gamma$ and $\bar\gamma$ lie in different path-components of $\operatorname{Imm}(S^1,\mathbb{R}^2)$. The figure-eight $(\sin 2t,\,\sin t)$ has $\tau=0$.
Step through five curves. On the left, $\gamma$ and its unit tangent $\mathbf{T}$. On the right, the direction of $\mathbf{T}$ is plotted as a polar angle with time as the radius, so that the path of $\mathbf{T}$ unrolls into a spiral instead of retracing the same circle. Count how many times the spiral goes around the centre. That integer is the obstruction. Written as a map $\mathbf{T}:S^1\to S^1$, it is the degree of $\mathbf{T}$, the element of $\pi_{1}(S^1)=\mathbb{Z}$ from Vol. II.
Right panel: the orange trace sits at polar angle equal to the direction of $\mathbf{T}(t)$ and at radius growing with $t$, so windings stack up instead of overlapping. The number of times it goes around the centre, counted with sign, is $\tau$.
Whitney–Graustein
Why $\tau$ cannot change. Along a regular homotopy $\gamma_{s}$ the velocity $\gamma_{s}’(t)$ is never zero, so $\mathbf{T}_{s}=\gamma_{s}’/\lVert\gamma_{s}’\rVert$ is defined and depends continuously on $(s,t)$. By the homotopy lifting of Vol. II, its angle $\theta_{s}(t)$ can be chosen continuous in $(s,t)$, so $\tau(\gamma_{s})$ is a continuous function of $s$ with values in $\mathbb{Z}$. By Vol. I, a continuous map from the connected space $[0,1]$ to the discrete space $\mathbb{Z}$ is constant. That is the whole proof, and it uses regularity exactly once: to divide by $\lVert\gamma_{s}’\rVert$.
Drop regularity and the argument breaks, visibly. The family below is a perfectly good homotopy. At one instant its velocity vanishes, and at that instant, and only then, $\tau$ jumps.
Drag $a$ across $\tfrac12$. The white dot is $\gamma_{a}(\pi)$ and the gold arrow is the velocity there, $(0,\,2a-1)$. It shrinks, vanishes at $a=\tfrac12$ (a cusp: the curve is not an immersion), and reappears pointing the other way. On the right, $\tau$ jumps from $1$ to $2$ at exactly that instant and at no other.
The computation behind the picture: with $z=e^{it}$, $\gamma_{a}=z+az^2$ and $\gamma_{a}’=iz\,(1+2az)$. The factor $iz$ turns once. The factor $1+2az$ winds around $0$ once if $2a>1$ and not at all if $2a<1$. So $\tau=1$ for $a<\tfrac12$, $\tau=2$ for $a>\tfrac12$, and at $a=\tfrac12$ the factor vanishes at $z=-1$, which is $t=\pi$.
Why equal $\tau$ is enough. Let $\gamma_{0}$ and $\gamma_{1}$ have the same turning number $\tau$. Rescale both to length $2\pi$ and parametrize by arc length (both steps are regular homotopies), so $\gamma_{j}’(t)=e^{i\theta_{j}(t)}$ with $\theta_{j}(2\pi)-\theta_{j}(0)=2\pi\tau$ for $j=0,1$. Interpolate the angles, not the curves: $\theta_{s}=(1-s)\theta_{0}+s\theta_{1}$. Integrating $e^{i\theta_{s}}$ need not give a closed curve, so subtract the average:
\[\gamma_s(t)=\gamma_s(0)+\int_0^t\bigl(e^{i\theta_s(u)}-c_s\bigr)\,du,\qquad c_s=\frac{1}{2\pi}\int_0^{2\pi}e^{i\theta_s(u)}\,du.\]Here $\gamma_{s}(0)=(1-s)\gamma_{0}(0)+s\gamma_{1}(0)$, and $\gamma_{s}(2\pi)=\gamma_{s}(0)$ because the integrand averages to zero. The velocity $e^{i\theta_{s}}-c_{s}$ is nonzero as long as $\lvert c_{s}\rvert<1$, and an average of unit vectors has length $1$ only if they all point the same way. When $\tau\neq0$ the angle $\theta_{s}$ must move, since $\theta_{s}(2\pi)-\theta_{s}(0)=2\pi\tau$ for every $s$, so they do not. (When $\tau=0$ one perturbs first to avoid a constant $\theta_{s}$.) At $s=0$ and $s=1$, $c_{s}=0$ because the original curves close up, so the family starts at $\gamma_{0}$ and ends at $\gamma_{1}$.
So the circle cannot be everted in the plane. Turning it inside out reverses the direction of travel, which sends $\tau=+1$ to $\tau=-1$, and no regular homotopy connects them.
The surprise is dimensional. The turning number of a curve is an element of $\pi_{1}(S^1)=\mathbb{Z}$. For a surface in $\mathbb{R}^3$ the analogous invariant lives in $\pi_{2}(V_{3,2})$, where $V_{3,2}$ is the space of linearly independent pairs of vectors in $\mathbb{R}^3$. That group is $0$ (Vol. V). One extra dimension is enough room for the obstruction to die.
The Gauss map is not the obstruction
Given an immersion $f$,
\[\mathbf{n}_f=\frac{\mathbf{f}_u\times\mathbf{f}_v}{\lVert\mathbf{f}_u\times\mathbf{f}_v\rVert}:S^2\to S^2.\]The degree of a map $g:S^2\to S^2$ counts signed coverings. Pick a target point $y$ such that $g$ is a local diffeomorphism near each of its preimages, and count $+1$ for each preimage where $g$ preserves orientation and $-1$ where it reverses it. The identity has degree $1$: one preimage, orientation kept. The antipodal map $p\mapsto-p$ has degree $-1$: one preimage, and it turns the outward side of the sphere inward. Wrapping the sphere twice around its axis, $(\theta,\varphi)\mapsto(\theta,2\varphi)$, has degree $2$. Hopf’s theorem says the count does not depend on $y$ and does not change under homotopy, which makes degree the integer of $\pi_{2}(S^2)\cong\mathbb{Z}$ from Vol. II. We will not need that theorem, because the Gauss maps of the two ends of an eversion turn out to be literally the same map. For the inclusion, $\mathbf{n}_{\iota}(p)=p$. For the everted sphere, $\alpha_{\theta}\times\alpha_{\varphi}=(-\mathbf{r}_{\theta})\times(-\mathbf{r}_{\varphi})=\mathbf{r}_{\theta}\times\mathbf{r}_{\varphi}$, so $\mathbf{n}_{\alpha}(p)=p$ as well. Whatever can be read off the Gauss map, degree included, is identical at the two ends. The Gauss map does not forbid eversion. (Readers who know Gauss–Bonnet can say more. With $K$ the Gauss curvature, the product of the two principal curvatures (the most and the least the surface bends, over all directions at that point), and $\chi$ the Euler characteristic, $V-E+F$ for any way of cutting the surface into polygons, which is $2$ for the sphere, the total curvature $\int K\,dA$ equals both $4\pi\deg\mathbf{n}$ and $2\pi\chi(S^2)=4\pi$. So every immersion $S^2\looparrowright\mathbb{R}^3$ has Gauss degree $1$.)
The canvas is an ellipsoid, not a round sphere, so that $\mathbf{n}$ is not the identity. For
\[\mathbf{f}(\theta,\varphi)=(a\sin\theta\cos\varphi,\;b\sin\theta\sin\varphi,\;c\cos\theta)\]the Gauss map is the normalization of $(x/a^2,\,y/b^2,\,z/c^2)$. The ellipsoid is strictly convex, so its Gauss map is a bijection onto $S^2$: for each unit vector $\mathbf{u}$ exactly one point of the ellipsoid sticks out farthest in the direction $\mathbf{u}$ (it maximizes $\mathbf{u}\cdot\mathbf{x}$), and the outward normal there is $\mathbf{u}$. It covers the target exactly once, degree $1$. The orange point on the right follows the unit normal as the white point tours the ellipsoid.
right: nf on the unit sphere
$\mathbf{n}_{f}=(\mathbf{f}_{\theta}\times\mathbf{f}_{\varphi})/\lVert\cdot\rVert$. White on the left is a point of the surface and the gold arrow its unit normal; orange on the right is the same unit vector, placed on the target sphere. The image of $\mathbf{n}_{f}$ is the whole target sphere, once.
The Gauss map remembers only the normal $\mathbf{n}=(\mathbf{f}_{u}\times\mathbf{f}_{v})/\lVert\cdot\rVert$, and its degree is the same for every immersed sphere. The circle case suggests the right move: in the plane the invariant was the direction of the whole derivative $\gamma’$ (for a curve the normal is just $\mathbf{T}$ turned through a right angle, so nothing is lost). For surfaces the whole derivative is a pair of vectors, a 2-frame, and the normal records only the oriented plane the pair spans. Why the space of such pairs, $V_{3,2}$, has no room for an obstruction is the next volume.