homotopic maps from the sphere to the sphere

i have some problems with the exercise 2.2.6 in hatcher book "algebraic topology". Hope that someone could help me out. One has to show that every map $S^{n} \rightarrow S^{n}$ can be homotoped to a map with a fixed point. Hope this is not too trivial. Thanks in advance.

beno

Solution 1:

One thing that is proved in that section is that any map without fixed points can be homotoped to the antipodal map. So it remains only to show that the antipodal map is homotopic to a map with a fixed point. A particular example of a homotopy is composition with a family of rotations. So the antipodal map is homotopic to its composition with a half rotation through some axis. Now does this map have any fixed points?

Solution 2:

This might be long but it's an explicit way to prove this . When f has a fixed point there is nothing to show . Consider the case when f does not have a fixed point . In this case f is homotopic to the antipodal map . Now we consider two cases . First when n is odd . When n is odd the antipodal map is homotopic to the identity map. So we need to worry about the case when n is even i.e. n =2m (say). Now consider the following homotopy

$H^{'}(x,t) : S^{2m} \times [0,\pi] \rightarrow S^{2m}$

$((x_{1},x_{2},...,x_{2m+1}),t) \mapsto (x_{1}Cost-x_{2}Sint,x_{2}Cost+x_{1}Sint,...,x_{2m-1}Cost-x_{2m}Sint,x_{2m}Cost+x_{2m-1}Sint , -x_{2m+1})$

So see that this map is well defined i.e. the image of a point lies in $S^{2m}$ . See that at $t = \pi$ the map is the antipodal map and at t = 0 the map is g ($\textit{say}$) i.e. $g : S^{2m} \rightarrow S^{2m}$ where

$ g(x_{1},x_{2},...,x_{2m+1}) = (x_{1},x_{2},...,-x_{2m+1}) $

So f is homotopic to the antipodal map and further when n is even the antipodal map is homotopic to the map g . So $ f \sim g $ . But see that g fixes all the points $ (x_{1},x_{2},...,x_{2m+1}) $ on $S^{2m}$ when $ x_{2m+1} = 0$ . So f is homotopic to a map that fixes point .