Re: S^n is simply connected, n>=2

Li Yi wrote:
Thank you. I have finally understand the sketch of the proof. But how
to prove that q doesn't hit all points of S^2?

W. Dale Hall wrote:

That's correct. However, there do exist continuous maps from S^1 onto
S^2 (that is, maps that pass through all the points on S^2). The purpose
of the exercise is to show that any map (such as one of these surjective
maps) can be homotoped to one that doesn't hit all the points of S^2.
Once that's done, then you can retract to a point.

Dale.



Let's recall the original problem:

Take S^n be the unit sphere surface of R^{n+1}.
For any loop p in S^n, find a loop q in R^{n+1}
such that

(1) p and q have the same basepoint;
(2) q consists of finitely many line segments;
(3) ||p(s)-q(s)||<=1 for all 0<=s<=1,

from which deduce that S^n is simply connected
for n>=2.


(I note with surpise that I had forgotten that we were talking
about any n>=2, rather than S^2 alone ... not that it matters
for the proof. I'll return to the usage of S^n rather than S^2
and apologize for my lapse of memory)

Note that q consists of finitely many line segments (projected
radially to the sphere S^n, of course). Thus, q consists of
the union of a finite number of geodesic (i.e., great-circle,
in the case n=2) arcs:

q = A_1 union A_2 union ... union A_k

The complement of q in S^n then consists of the intersection:

(S^n \ A_1) intersect (S^n \ A_2) intersect ... intersect (S^n \ A_k).

Each of the sets S^n \ A_j is open and dense in S^n, which is a complete
metric space. The Baire category theorem says that a countable
intersection of such sets is nonempty, so we're done.

Dale
.

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