Authentic Personal message at 18:32:43 on Wed Mar 18 1992
From: Be there -- aloha! <zeno> on MADMAN.MIT.EDU
To: mosquito@ATHENA.MIT.EDU
Hmmm, maybe you could answer this one easily, then.  I wasn't able to get
it, but I didn't really think about it too long. 
If X is a nonempty compact set in the plane R^2, then it's easy to show that
there is some square in the plane which "circumscribes" X -- i.e. no point of X
lies outside the square, and each of the square's four sides touches X.  (If X
is a line segment, we have the degenerate case in which the square is placed so
that X is a diagonal).  To see this, just consider a given angle \theta and look
at all the lines in the plane forming that angle with the x-axis.  By compact-
ness, this family of lines will have one which just touches X "on one side" and
another, parallel one, just touching X "on the other side".  Let f(\theta) = the
perpendicular distance between these two lines.  [This is the same as the
diameter of X's projection onto the line through the origin at angle 
\theta + \pi/2 ].  So  f:R/(2\pi Z) --> R  is continuous.  Then the function
F := f(\theta) - f(\theta + \pi/2)  is also continuous, and clearly must change
sign as \theta goes through an interval of length \pi.  So  F(\theta) = 0  for
some \theta, and for this value, the pair of lines at angle \theta which bound X
are the same distance apart as the bounding pair at angle \pi/2 from \theta, so
at this orientation there's a square circumscribing X.  
NOW, if that made sense ... consider the same problem for compact X in R^3.
There is in fact a circumscribing cube, but no simple generalization of the
above proof suggests itself.  I am told, however, that it's "straightforward" if
you use the fact that the fundamental group of S^3 is Z/2Z.  See any way to
handle this?  It might help to use the realization of S^3 as a solid ball in R^3
with each surface point identified with its antipodal point.  So you can
identify cube orientations (let's say always origin-centered) with points of S^3
in a way that would take me too long to outline via keyboard.
Actually, I guess .906 mostly deals with {,co}homology, not homotopy.  Oh well.
Authentic Personal message at 18:48:59 on Wed Mar 18 1992
From: Be there -- aloha! <zeno> on MADMAN.MIT.EDU
To: mosquito@ATHENA.MIT.EDU
I don't remember the way to demonstrate that this isn't a null-homotopic
loop, but using that realization I described before (solid ball in R^3
with antipodal surface points identified), I know that the loop starting
at the center point, going straight out along some axis, hitting the
"surface" and continuing back in on the other side of the same axis and
back to the center, is such a loop.  There's a great series of diagrams
in vol. 2 of Berger's "Modern Geometry", in the chapter on spheres,
which makes this all intuitive.  It's related to a certain parlor
trick involving holding a bowl of soup in your hand and rotation that
arm through a full vertical circle without spilling any soup.