Antipodal map of spheres

Question

Let $A: S^2 \to S^2$ be the antipodal map between spheres, we know that it induces the endomorphism $-1$ on the $\pi_2(S^2)=Z$. Consider the composite $S^2 \xrightarrow{A} S^2 \xrightarrow{\pi} \mathbb{R}P^2$, the second arrow being the quotient map. Now this map is the same as the projection map $\pi:S^2 \to \mathbb{R}P^2$, and the projection map being a covering space map induces an isomorphism between $\pi_2(S^2)$ and $\pi_2(\mathbb{R}P^2)$. Now, it follows that $\pi_* = A_* \pi_* = -\pi_*$ as maps $\pi_2(S^2) \to \pi_2(\mathbb{R}P^2)$. Isn't that a contradiction? Or am I missing something here?

Answer 1

Note that $A$ is not basepoint-preserving. This way, you can not just apply the functor $\pi_*$ to the commuting triangle $S^2 \xrightarrow{A} S^2 \to P^2.$ Indeed, recall why the projection gives an isomorphism: having $S^n \to P^2$ you choose a lift $S^n \to S^2$ by lifting the map piece by piece on trivializing neighborhoods, starting from the neighborhood of the basepoint. Without the basepoint, you'd have two choices of the first step and will potentially have two listings - that's not a bijection.