Hatcher Exercise 2.2.3
Theorem: If is of degree zero, then there are points with and .
Proof: If does not have a fixed point, then it is of degree . Conversely, if is of degree zero it must have a fixed point.
Since is of degree zero, we have as well since degree is multiplicative with respect to composition. Then, by the same argument as before, we conclude that must have a fixed point , so that and thus .
Corollary: A continuous non-vanishing vector field on has a point on the boundary where points radially inward, and another where it points radially outward.
Proof: Since is non-vanishing, we can normalize it, letting . Then, . Now, we have a map . Since the map factors through a contractible space, , and is of degree zero. Then, applying the theorem we have and with and . Then, since the first map in is just the inclusion, we have and . Then, for the original map , we conclude is parallel to and is opposite in direction to . Since and point in the radial direction, this means that “points outward” and “points inward”.