# non-homotopic maps $\mathbb S^2\rightarrow \mathbb{P}_\mathbb{R}^2$

How can I find non-homotopic maps $\mathbb S^2 \rightarrow \mathbb{P}_\mathbb{R}^2$? I know that it is enought that the degree are different. But the canonical map given by the antipodal map has degree $0$ and I can't find another map of degree differente than zero.

• What degree do you have in mind? Since the codomain of your maps is not orientable, you need to use $Z/2Z$ coefficients and that is not going to be of much use. – Mariano Suárez-Álvarez Jun 12 '16 at 21:05
• The covering map $\pi:S^2 \to \mathbb{RP}^2$ induces isomorphisms $\pi_n S^2 \to \pi_n \mathbb{RP}^2$ for $n > 1$. (Think of $\pi$ as a fibration with discrete cover.) – anomaly Jun 12 '16 at 21:24

Using degrees to disprove homotopicity is difficult in this case. However, if there were a null-homotopy of the canonical map, you can lift that to a null-homotopy of the identity map on $S^2$.
As is the case for all covering maps, the canonical map $p:S^2\to \mathbb P^2_\mathbb R$ induces an abelian group isomorphism $$\pi_2(p):\pi_2(S^2)\stackrel {\cong }{\to} \pi_2 (\mathbb P^2_\mathbb R):[Id]\mapsto [p]$$ This proves that the the group $\pi_2 (\mathbb P^2_\mathbb R)$ is isomorphic to $\mathbb Z$, a generator being the class $[p]\in \pi_2 (\mathbb P^2_\mathbb R)$.
Now if $\:g_n:S^2\to S^2 \; (n\in \mathbb Z)\:$ is a map of degree $n$, it will induce multiplication by $n$ on $\pi_2(S^2)$, i.e. $$\pi_2(g_n):\pi_2(S^2)\to \pi_2(S^2):[Id]\mapsto n[Id]$$ Finally the map $p\circ g_n$ induces in homotopy $$\pi_2(p\circ g_n)=\pi_2(p)\circ \pi_2(g_n):\pi_2(S^2)\to \pi_2 (\mathbb P^2_\mathbb R):[Id]\mapsto n[p]$$ Since these maps are different for different $n$'s, we obtain a family of maps $p\circ g_n:S^2\to \mathbb P_\mathbb R$ ($n\in \mathbb Z$) no two of which are homotopic, just as required.
But how do I obtain a map $\textbf {g}_\textbf {n}$ of degree $\textbf{n}$ ?
Very easily! Interpret $S^2$ as the extended complex plane $\hat{ \mathbb C}$ and take for $g_n$ the map $g_n(z)=z^n$.