"Let f,g:S(adsbygoogle = window.adsbygoogle || []).push({}); ^{n}→S^{n}be maps so that f(x) and g(x) are not antipodal for any x. Show that f and g are homotopic."

Here's my initial approach: I figured it would be easier to work in I^{n}instead, so I note that S^{n}is the quotient of the n-cube with its boundary. Therefore, each map S^{n}→S^{n}can be lifted to a map I^{n}→S^{n}which is constant on the boundary in R^{n}.

It may also be possible to use a covering space to lift the other end of the map, getting I^{n}→R^{n}. I played around with this in the n=1 case (the only case where I'm certain it'd work) and it didn't seem to help.

In any case, I^{n}is contractible, so we have a homotopy F between f' and g' (our lifted maps). I just can't figure out how to use the fact that f and g are never antipodal to prove that F can be deformed into a homotopy which is constant on the boundary at ever stage. I understand intuitively that that restriction prevents one map from ever "wrapping around" more than the other, but I'm not certain how to apply it.

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# Homotopic maps of the n-sphere

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