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Let $\phi:D^3\to S^2$ be the composition $D^3\to S^3\to S^2$, the first map being the quotient by the boundary and the second map being the Hopf map. Then: $$f_t:x\mapsto(1-t)x+t\phi(x)$$ is a homotopy that takes every point in $D^3$ on a straight-line path towards its destination in $S^2$. Furthermore, everything on the boundary travels to the same point.

Can anyone help me visualize that homotopy? Preferably with an animation; I must confess that I have no idea how to make one myself.

Also, if there's another homotopy that's easier to visualize, where the points travel from $x$ to $\phi(x)$ but not necessarily on a straight-line path, that would be helpful too.

I apologize if this question isn't suitable for this site.

Akiva Weinberger
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  • On the surface it's a reasonable question, but I think you're unlikely to find the result satisfying: (i) The entire boundary of the ball maps to one point in $S^{3}$, and (ii) The quotient $S^{2}$ doesn't sit inside $S^{3}$ in a nice way. (Particularly, the image of the Hopf map is not an equatorial sphere; it's more like an equatorial hemisphere with the boundary circle collapsed to a point.) Is there some aspect of the Hopf map you're trying to understand? Have you done a web search for images of the Hopf map, and do the results (not) make sense? – Andrew D. Hwang Jun 08 '16 at 01:46
  • I think the result might be interesting precisely because the image doesn't sit inside $S^3$ nicely. How do the domain and image interact, then? I found the visualizations online to be quite nice and pretty and clear, but there's a certain discreetness about them; the domain is over here, the image is over there, and there's nothing in between. I feel like the Hopf map must either be like "smooshing a ball onto the surface of a sphere", or fail to be that in some interesting way. – Akiva Weinberger Jun 08 '16 at 01:55
  • Also, a more specific question: Does the homotopy pass through every point inside the sphere? Can we do any homotopy like this in $\Bbb R^3\setminus{0}$? (Conjecture: No.) [EDIT: You'd need to shift over $D^3$ in $\Bbb R^3$ so that it doesn't contain $0$.] I think the animation could help answer questions like that. – Akiva Weinberger Jun 08 '16 at 01:56
  • In the hope it helps, the diagram in my answer to Flag manifold to Complex Projective line is the best picture I know; if you really understand that analogy with $\mathbf{C}^{2}$, you can see that the intersections of complex lines with $S^{3}$ are equatorial circles, and shrinking them so their radius decreases to zero has exactly the effect of smooshing $S^{3}$ onto $S^{2}$, in the same way that shrinking the width of a Moebius strip to zero sends the boundary circle to the central circle as a two-to-one covering. [...] – Andrew D. Hwang Jun 08 '16 at 02:15
  • Your proposed homotopy joins the identity map of $D^{3}$ and some mapping of $D^{3}$ to $S^{2}$ followed by a (non-natural) inclusion of $S^{2}$ back into $D^{3}$. In that sense, yes, it passes through every point of $D^{3}$ (and $S^{3}$, and $S^{2}$). – Andrew D. Hwang Jun 08 '16 at 02:20
  • To flesh out the Moebius strip comment: The total space of $\mathbf{C}^{2}$ blown up at the origin is the complex analog of a Moebius strip (the total space of the tautological bundle over a projective line). The $3$-sphere may be viewed as the set of vectors of norm $r > 0$; as $r \searrow 0$, the sphere collapses (i.e., smooshes) onto the complex projective line by the Hopf map. – Andrew D. Hwang Jun 08 '16 at 02:26

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