If $f : S^1 \to S^1$ is odd, then $f$ cannot be nullhomotopic The problem is:
Let $f:S^1 \to S^1$ be a continuous map that has the property that $f(-z) = -f(z)$ and $f(1) = 1$. Prove $f$ cannot be null-homotopic.
Any ideas? The first step is to show that by defining a map $p: S^1 \to S^1$, $p(z) = z^2$, there is a map $g:S^1 \to  S^1$ so that $g$ and $f$ commute.
 A: Since the proof of the statement "odd maps $S^n\to S^n$ have odd degrees" is given in Hatcher's textbook on page 174 - 176. The proof uses in particular the exact sequence for the covering space $p:S^n\to\mathbb{R}P^n $ with coefficients in $\mathbb {Z}/2\mathbb{Z} $, i.e.
$$...\to 0 \to H_n (\mathbb{R}P^n,\mathbb {Z}/2\mathbb{Z})\overset{\tau_\ast}{\to}H_n (S^n, \mathbb {Z}/2\mathbb {Z})\overset {p_\ast}{\to} H_n (\mathbb{R}P^n,\mathbb {Z}/2\mathbb{Z}) \to H_{n-1}(\mathbb{R}P^n,\mathbb {Z}/2\mathbb{Z}) \to 0 \to... $$
This is associated with the short exact sequence 
$$0\to C_n(\mathbb{R}P^n,\mathbb {Z}/2\mathbb{Z}) \overset {\tau}{\to}C_n (S^n,\mathbb {Z}/2\mathbb{Z})\overset {p_c}{\to}C_n (\mathbb{R}P^n,\mathbb {Z}/2\mathbb{Z})\to0$$
where the first map $\tau_\ast:H_n (\mathbb{R}P^n,\mathbb {Z}/2\mathbb{Z})\to H_n (S^n, \mathbb {Z}/2\mathbb {Z})$ is induced by the map $\tau$ which sends each singular simplex $\sigma:\Delta^n\to \mathbb{R}P^n$ to the sum of its two lifts to $\bar{\Delta^n}$. By this definition $\tau $ is injective. The induced map $p_c$ on the chain groups is surjective and the kernel is generated by sums of the form $ \bar {\sigma}_1 +\bar {\sigma}_2$, since each $\sigma $ has exactly two lifts $\bar{\sigma}_1$ and $\bar {\sigma}_2$ and the coefficients are in $\mathbb {Z}/2\mathbb {Z} $.
We can also conclude that  $im (\tau)=\ker (p_c) $, hence we have the short exact sequence.
Then one can show that an odd map $f:S^n\to S^n $ induces a quotient map $\bar {f}:\mathbb {R}P^n\to \mathbb {R}P^n $ and these two maps induce maps on the chain groups, s.t. the diagram of the short exact sequence above commutes with the two induced maps.
The corresponding commutative diagram on the homology groups shows that all of the two induced maps $f_\ast $ and $\bar {f}_\ast $ on the homology groups are isomorphisms (induction! Start with dimension 0). 
Since the iso $f_\ast:H_n (S^n,\mathbb {Z}/2\mathbb {Z})\to H_n (S^n,\mathbb {Z}/2\mathbb {Z})$ is multiplication by $\deg f \mod 2$, the degree is odd.
Obviously the degree of a constant map $c $ is $0$, because if we choose an $x_0$ which is not in the image of the const. map, then it factors through $S^n\setminus x_0$ which is homeomorphic to the $\mathbb {R}^n $ by stereographic projection. The degree of homotopic maps is equal, since they induce the same maps on the homology groups. But $2k+1=\deg f=\deg c =0$ is a contradiction.
