How does one prove that any self homeomorphism of $\mathbb{C}P^2$ is orientation preserving. I think this question is not very difficult,but I don't solve it.
 A: Claim: If $f:\mathbb{CP}^2\to \mathbb{CP}^{2}$ is a homeomorphism, then $f$ is orientation preserving. 
Proof: You can do this via algebraic topology. The top homology group of $\mathbb{CP}^2$ is $H^4(\mathbb{CP}^2)=\mathbb{Z}$. 
Exercise 1: A homeomorphism $f:\mathbb{CP}^2\to \mathbb{CP}^2$ is orientation preserving if and only if the induced homomorphism $f^{4}:H^4(\mathbb{CP}^2)\to H^4(\mathbb{CP}^2)$ is the identity. (This might be tricky depending on your definition of "orientation preserving"; let me know if you have difficulties with it and I'm happy to help.)
Let's look at the induced homomorphism $f^{2}:H^2(\mathbb{CP}^2)\to H^2(\mathbb{CP}^2)$. Recall that $H^2(\mathbb{CP}^2)=\mathbb{Z}$ and that if $x\in H^2(\mathbb{CP}^2)$ is a generator, then $x^2$ is a generator of $H^4(\mathbb{CP}^2)$ (here I am referring to the cup product in singular cohomology). 
Exercise 2: Prove that $f^4$ is the identity by looking at $f^2$ and using the naturality of the cup product.
Exercises 1 and 2 imply that $f:\mathbb{CP}^2\to \mathbb{CP}^2$ is orientation preserving.
I hope this helps!
