I am working through 4-Manifolds and Kirby Calculus by Stipsicz and Gompf. At the beginning of Section 1.3, they have a list of exercises regarding $\mathbb{C}P^n$ and $\mathbb{R}P^n$. The part I couldn't figure out is as follows:
Prove that there is no orientation-preserving diffeomorphism between $\mathbb{C}P^2$ and $\overline{\mathbb{C}P^2}$.
Here $\overline{\mathbb{C}P^2}$ denotes $\mathbb{C}P^2$ with the opposite orientation. This question is asked right after showing that the intersection form of $\mathbb{C}P^2$ is $\langle 1 \rangle$ while the intersection form of $\overline{\mathbb{C}P^2}$ is $\langle -1 \rangle$. I am assuming I somehow need to make use of this fact. Here is an answer from another blog, but I couldn't really understand how it works.
Okay, I think I have an answer now. We know that if $a\in H^2(\mathbb{C}P^2) \cong \mathbb{Z}$ is a generator, then $a^2 = a\smile a\in H^4(\mathbb{C}P^2)$ is simply $[\mathbb{C}P^2]$. Consequently, $-a^2 = -a\smile a = [\overline{\mathbb{C}P^2}]$. Suppose we have a map $f:\mathbb{C}P^2\xrightarrow{} \mathbb{C}P^2$. Then $f_\ast(a) = ca$ for some $c\in \mathbb{Z}$, which in turn implies $$f_\ast(a\smile a) = f_\ast(a)\smile f_\ast(a) = ca\smile ca = c^2 (a\smile a) $$ Since $c^2 \neq -1$, there is no map $f$ that reverses the orientation.