Prove some facts about fixed points on the sphere using differential topology I'm reading Differential Topology by Guillemin & Pollack, and I would like to know how to prove these facts about fixed points on the sphere in this viewpoint:)
Let $f,g \in {C^0}({\mathbb{S}^{k}},{\mathbb{S}^{k}})$.
(1) If $\deg f \ne {( - 1)^{k + 1}}$, then there exists $x_0$ s.t. $f(x_0)=x_0$.
(2) If $\deg f \ne 1$, then there exists $x_0$ s.t. $f(x_0)=-x_0$.
(3) if $k$ is even, then one of the three maps $f,g,g \circ f$ has a fixed point.
I've been working on them for a while but still can't see the path. Maybe my understanding is not deep enough... Could anyone give me some instructions? Any help will be appreciated:)
 A: As said in the comments there are good approximation theorems to pass from the smooth to the continuous case, which is why I'll be considering smooth functions from here on out. 
A useful result when dealing with antipodes is the following :

Lemma 1 : Let $X$ be a smooth manifold and $f,g : X \to S^n$ smooth functions such that that for all $x \in X$,
  $$
||f(x)-g(x)||<2
$$
  Then $f$ and $g$ are smoothly homotopic. 

It is a good exercise to prove this (consider the straight line homotopy from $f$ to $g$. What can you say about it ?). The second observation you might need is that the antipodal map $a :S^n \to S^n$ 
$$
(x_1,...,x_{n+1}) \mapsto (-x_1,...,-x_{n+1})
$$
has degree $(-1)^{n+1}$. To show this rewrite it as a composition of reflexions on the $i^{th}$ coordinate, and use the fact that the degree of a composition is the product of the degrees.
Equipped with these results we can show what you want :
(1)  Consider the contraposition, and suppose that $f$ has no fixed point. Then, for all $x \in S^n$,
$$
||f(x)-(-x)||<2
$$
(You can show this). What does lemma 1 tell you ? Can you conclude ?
(2) You can use the exact same line of reasoning (remember that the identity map has degree 1).
(3) Suppose that none of these mappings have a fixed point. Then by (1), $f,g$ and $g \circ f$ all have degree $(-1)^{k+1}=-1$ (since $k$ is even). However, 
$$
deg(g\circ f)=deg(g)deg(f)=(-1)^2=1
$$
Which is a contradiction. 
A: (1) and (2) are equivalent since the antipodal map has degree $-(-1)^k$.
To prove (2), if $f(x)\ne-x$ then there is a homotopy from $f$
to the identity map, defined by following the shortest
path on the sphere from $f(x)$ to $x$.
For (3), note that if not, all of $f$, $g$ and $g\circ f$ would have degree $-1$.
