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Recently, I am learning Bertini's theorem because I encounter "generic smooth" problem during my research. I'm not an algebraic geometer and I read the Hartshorne Chapter 3 Theorem 10.8 to learn "generic smooth". Kleiman's result show that we can require $f^{-1}(l)$ to be smooth for generic 1-dimensional complex straight line $l$. But now, I want to have a stronger result, which requires $l$ passing through a fixed point $P$ not contained in the critical value. More precisely, I want to know whether the following question is true.

Suppose $f: \mathbb{CP}^n\to \mathbb{CP}^n$ is a surjective endomorphism. Let $R$ denote its critical value, which is a hypersurface on $\mathbb{CP}^n$. Fix a point $P\in \mathbb{CP}^n\setminus R$, and parametrize all 1-dimensional complex straight line $l$ passing through $P$ by $\mathbb{CP}^{n-1}$. We use $f^{-1}(l)$ to denote the scheme-theoretic preimage of $l$ in $\mathbb{CP}^n$. I want to prove the following result:

For generic such $l$ passing through $P$ (parametrized by $\mathbb{CP}^{n-1}$ at $P$), $f^{-1}(l)$ is a smooth curve (not necessarily irreducible) in $\mathbb{CP}^{n}$. Thus, $f|_{f^{-1}(l)}$ is also a ramified cover map between Riemann surface $f^{-1}(l)$ and $l$.

I can't apply Hartshorne's chapter 3 Theorem 10.8 directly. I draw some pictures in $\mathbb{CP}^2$ and feel it should be right. If it is wrong for an arbitrary point $P$, I guess that $P$ satisfying the above property should consists of a dense set in $\mathbb{CP}^{n}$. But I don't know how to show this. Could anyone give me some references about it? Or if it wrong, how to construct a counter-example?

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  • $\begingroup$ Oh, I think I know how to do. Consider the projection $ \mathbb{CP}^n\setminus f^{-1}(P)\to \mathbb{CP}^{n}\setminus P \to \mathbb{CP}^{n-1}$. Every point of $\mathbb{CP}^{n-1}$ corresponds with a complex straight line passing through $P$. Use Kleiman's result (Hartshorne chapter 3 Theorem 10.8) for $ f: pt \to \mathbb{CP}^{n-1}$ and $g: CP^{n}\setminus f^{-1}(P) \to \mathbb{CP}^{n-1}$. $\endgroup$
    – MATHQI
    Commented Dec 5 at 9:57
  • $\begingroup$ Re, don't forget that you can (and probably should) post an answer to your own question. $\endgroup$
    – LSpice
    Commented Dec 5 at 14:27

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