For a function with odd degree $f(x)$ is not homotopic to $f(-x)$ Let $f$ be a function from $S^n$ to $S^n$ which has odd degree. How can I show that $f(x)$ is not homotopic to $f(-x)$?
 A: This is not true. The antipodal is homotopic to the identity if $n$ is odd.
Antipodal map homotopic to identity for $S^{2n-1}$
A: This is false for $n$ odd. 
Claim 1: for $n$ odd, it is not the case that there is always some $x$ so that $f(x)=f(-x)$. take $n=1$ for example, consider the map $x \mapsto e^{i\pi/2}x$. This is a quarter rotation, so clearly there is no $x$ so that $x=-x$.
Claim 2: for $n$ odd, $x \mapsto f(-x)$ and $x \mapsto f(x)$ may be homotopic, even if the degree of $f$ is odd.
Take the identity map on $S^1$ and the antipodal map $x \mapsto -x$. Then consider the homotopy $(x,t) \mapsto e^{it}x$ for $t \in [0,\pi]$.

When $n$ is even, $f:S^{n} \to S^n$ induces a map $f_*:x \mapsto \alpha x$ on $H_n(S^n) \to H_n(S^n)$ which is called the degree of the map. 
write $h:S^n \to S^n$ for the antipodal map $x \mapsto -x$. $x \mapsto f(-x)$  is the composition  $(f \circ h)_*=f_* \circ h_*$ but the degree of $h$ is $(-1)^{n+1}$ since $h$ is the composition of $n+1$ reflections. 
So, $(f \circ h)_*=-f_*$ and they cannot be homotopic, or they would induce the same map on homology.
