Question about self homeomorphism of $\mathbb C\mathbb P^2$

Question

Can anyone give me any idea about how to show that: any self homeomorphism of $\mathbb C\mathbb P^2$ is orientation preserving? Thanks.

Answer 1

You can read this fact in the (co)homology groups with the cap & cup products. Let $f:\mathbb {CP}^2\rightarrow \mathbb {CP}^2$ be a self-homeomorphism and $[\mathbb {CP}^2]$ be the orientation of $\mathbb{CP}^2$ in $H_4(\mathbb {CP}^2)$.

If $a\in H^2(\mathbb {CP}^2)$ is a generator, then $f^\star(a)$ must be equal to $\pm a$ since $f^\star$ is an isomorphism.

On the other side, $a^2\in H^4(\mathbb {CP}^2)$ is also a generator or if you prefer, the bracket $<a^2,[\mathbb {CP}^2]>$ is equal to $1\in H_0(\mathbb{CP}^2)$.

Now, assume that $f$ reverses the orientation, i.e. $f_\star([\mathbb {CP}^2])=-[\mathbb {CP}^2]$ and play with the bracket $<a^2,[\mathbb {CP}^2]>=1$ in order to get a contradiction.