Projective plane and homotopies If we consider $P^2$ (the real projective plane) like a disk with its boundary identified via the antipodal map, we can consider, in this model, the 180 degrees rotation. It's easy to see that this rotation and the identity are free-homotopic. Are they homotopic if we consider pointed homotopies?
 A: Nice question.
Think about the disk as $D^2 \subset \mathbb R^2 \times \{0\} \subset \mathbb R^3$. Let the base point be $A = (1,0,0) \in \partial D^2$, with antipodal point $-a = (-1,0,0)$. 
Also think about the unit sphere $S^2 \to \mathbb R^3$, and its quotient map $q : S^2 \to P^2$ under which antipodal points on $S^2$ are identified. Let $p$ also be the base point of $S^2$. Let $a = q(A)$ be the base point of $P^2$.
The map $q$ is a universal covering map, and like all universal covering maps it induces an isomorphism $q : \pi_2(S^2,P) \to \pi_2(P^2,p)$. 
Let $\rho : P^2 \to P^2$ be your $180^\circ$ rotational map. That map may be lifted to a $180^\circ$ rotational map $\tilde\rho : S^2 \to S^2$. But, that map takes $P$ to $-P$. Composing with the antipodal reflection $\alpha : S^2 \to S^2$, the map $\alpha \circ \tilde\rho : S^2 \to S^2$ fixes $P$ and is a lift of $\rho$. Moreover, $\alpha \circ \tilde\rho$ induces the map on $\pi_2(S^2,Q) \approx \mathbb Z$ that swaps the two generators.
Therefore, the induced map of $\rho$ on $\pi_2(P^2,q) \approx \mathbb Z$ is the map that swaps generators, therefore $\rho$ is not homotopic to the identity rel base point.
