# Milnor immersion of circle, disks, and a ball

Milnor surprisingly found an immersion of a circle in the plane which bounds two "incompatible" immersed disks (see [1] for example, picture included). I think I can glue these two disks together to define an immersion of a sphere $$f:S^2\to\mathbb{R}^3$$ where Milnor's immersed circle (the equator) sits in the $$xy$$-plane and contracts to a point (in the $$\pm z$$ directions) by running along either immersed disk.

Does $$f(S^2)$$ bound an immersed ball, i.e. does $$f$$ extend to an immersion of a 3-ball?

This question was sparked by glancing at Gromov's book [1] and a related paper of Eliashberg-Mishachev which argues that the projection $$S^2\to\mathbb{R}^2$$ (composing $$f$$ above with projection onto $$xy$$-plane) is not homotopic through folded maps to the "standard" folded map $$(x,y,z)\mapsto (x,y)$$ on the unit 2-sphere (the fold being the equator).

If one assumes a couple of simple conditions on $$f$$, namely that the restriction of $$z$$ on $$f(S^2)$$ is a Morse function with two critical points and that projections of two halves of $$f(S^2)$$ ($$z>0$$, $$z<0$$) onto the plane $$z=0$$ realize two original different immersions then $$f$$ can not be extended to an immersion of the ball.

Indeed, assume by contradiction that such an immersion $$F:B^3\to \mathbb R^3$$ exists. Let $$m<0$$ be the minimum of $$z$$ on $$f(S^2)$$ and $$M>0$$ be the max. In this case the preimages of planes $$z=c$$ under $$F$$ will be $$2$$-disks in $$B^3$$ for all $$c\in (m,M)$$. And by continuity, (changing $$c$$ from $$M$$ to 0, or from $$m$$ to $$0$$) we will get that the map $$F$$ applied to the disk $$F^{-1}(z=0)$$ realizes both immersions, which is absurd.

One can relax the two conditions by requiring that there is a regular function $$z'$$ on $$\mathbb R^3$$ such that $$z'=0$$ is the plane $$z=0$$ and $$z'$$ restricts to $$f(S^2)$$ as a Morse function with two critical points. If I understand correctly this condition is implicit in your construction of $$f$$. (One should express what it means that the two halves of $$S^2$$ realize two different immersions of a disk.)

PS. To clarify the above reasoning I'll give a proof in the most restrictive case. This case already contains the main idea. So the setting will be the following

Setting. Let $$S^2$$ be the unit $$2$$-sphere, $$S^1\subset S^2$$ be the equator, $$O_+$$ be the north pole and $$O_-$$ the south. Let $$S^2_+$$ be the open upper half-sphere and $$S^2_-$$ the lower one. Let us assume that the immersion $$f: S^2\mathbb\to \mathbb R^3$$ has the following properties.

1) The restriction of $$f$$ to the equator $$S^1\subset S^2$$ realizes the Milnor circle.

2) Function $$z$$ restricts to $$f(S^2)$$ as a Morse function with two critical points so that $$\max_z f(S^2)=1$$, $$\min_z f(S^2)=-1$$ and moreover the poles are sent to critical points $$z(f(O_+))=1$$, $$z(f(O_-))=-1$$.

3) Let $$\pi: (x,y,z)\to (x,y)$$ be the projection. Then the maps $$\pi\circ f: S^2_+\to \mathbb R^2$$ and $$\pi\circ f: S^2_-\to \mathbb R^2$$ realize the two Milnor immersions.

So, suppose all these conditions hold. Then each set $$F^{-1}(z=c)$$ is a disk $$D_c\subset B^3$$ for $$c\in (-1,1)$$. Consider the family of immersed disks $$\pi(F(D_c))\subset \mathbb R^2$$ for $$c$$ varying from $$1$$ to $$0$$. I claim that this family of immersed disks is the same as the following second family of immersed disks:

Second family. Let $$D_c^+\subset S_+^2$$ be the disk composed of points with $$z\ge c$$. Then we can consider the family of immersed disks $$\pi(F(D_c^+))$$.

So the claim is that the two families $$\pi(F(D_c))$$ and $$\pi(F(D_c^+))$$ coincide.

But then, the immersion $$\pi(F(D_0))$$ is the same as $$\pi(F(D_0^+))=\pi(F(S_+^2))$$.

Just in the same way we prove that $$\pi(F(D_0))$$ is the same as $$\pi(F(S_-^2))$$. This contradicts 3).