More with fibers of functions

I posted earlier on a way of visualizing the fibers of certain maps from high dimensions to low dimensions.  Specifically, if the range can be embedded in the domain so that f is the identity of the image of the range, then we can draw the inverse image at each point.  I had some images of functions whose inverse image was a torus, but had trouble making these sorts of images for maps f: \Omega \subset \mathbb{R}^3 \to \mathbb{R}^2, so that the inverse image of a point is a line.  Well, no more!  Here are two images, one is the projection of a cube onto a square, and the other is somewhat more complicated, and is the string hyperboloid map.  See the previous post for more details on these specific maps, but I just thought these were nice images!

Fibers of the projection map from the cube to the square.

Fibers of the "twisted cylinder", which are again straight lines.


3 comments on “More with fibers of functions

  1. Leonid Kovalev says:

    In the first case the fibers are equidistant; are they in the second? Two sets A, B are equidistant if for every a\in A there is b\in B such that d(a,b)=dist(A,B); and vice versa.

    • Leonid Kovalev says:

      That was a before-morning-coffee question… of course not, since they are straight lines. Two line segments can’t be equidistant unless they are parallel.

  2. Yes, but even in a colloquial sense, they are not equidistant: there are 9 radii, with something like 20 fibers per radius. So the fibers on the smallest hyperboloid are much more tightly bunched than the outer hyperboloid.

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