"Let f,g:Sn→Sn 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 In instead, so I note that Sn is the quotient of the n-cube with its boundary. Therefore, each map Sn→Sn can be lifted to a map In→Sn which is constant on the boundary in Rn. It may also be possible to use a covering space to lift the other end of the map, getting In→Rn. 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, In 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.