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I think this question is not very difficult,but I don't solve it.

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Can you show any work of your own? What have you tried, where have you failed, etc. –  user641 Jul 7 '13 at 14:57
Homeomorphism as a topological space? Manifold? Complex manifold? –  Jyrki Lahtonen Jul 7 '13 at 15:40
It seems as a complex manifold, because it looks like there is probably an orientation reversing automorphism as a manifold. –  Matt Jul 7 '13 at 16:13
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1 Answer

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!

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