mapping class group of the real projective plane In most literature I've read about the mapping class group, I found that many authors have stated without any explanation that any homeomorphism of a real projective 2-space to itself is isotopic to the identity. I'm guessing it is obvious but I can't seems to come up with a sound explanation, this is what i have:
$\mathbb{R}P^2$ can be constructed from from glueing the boundary of a disk $D^2$ to the boundary of a Mobius band, since the mapping class group of D^2 and the Mobius band are both trivial then the mapping class group of $\mathbb{R}P^2$ is trivial. 
the reason why this doesn't seem sound is because the Klein bottle can be constructed from glueing two Mobius band's together by their boundary but the mapping class group of the Klein bottle is not trivial. 

The free mapping class group (that is, the path components of its full group of homeomorphisms) of the Möbius band M is not trivial. It has a self-homeomorphism h : M → M whose effect h* : H1(M;Z) →  H1(M;Z) on the first homology group (which is infinite cyclic) is multiplication by -1.
The projective plane P2 can be modeled as the 2-disk D2 with each pair of antipodal points on its boundary S1 identified. From this model it's easy to see that there is an isotopy of the identity id : P2 → P2 to a homeomorphism taking the generator of H1(P2;Z) (as a simple closed curve) to itself going in the reverse direction. (Just rotate the 2-disk by 180º.) This shows that the inclusion of a Möbius band into P2 can be reversed by an isotopy within P2.
