Let $X$ be a projective variety. Why is any surjective morphism from $X$ to itself finite?
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asked Aug 7, 2014 at 9:38
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$\begingroup$ I've already asked this question on MSE (math.stackexchange.com/questions/885733/…) deeming it a not so hard one, but I haven't received enough hints to solve it. Please answer it there if you find it more suitable for MSE. $\endgroup$– Davide Cesare VenianiAug 7, 2014 at 9:39
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$\begingroup$ This sounds like Ax-Grothendieck. $\endgroup$– Jason StarrAug 7, 2014 at 10:19
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1 Answer
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Let $f$ be your surjective endomorphism; it is generically finite, say of degree $d$. Then $f_*:H^*(X,\mathbb{Q})\rightarrow H^*(X,\mathbb{Q})$ is surjective, because $\ f_*f^*=d.\mathrm{Id}\ $ (use $\mathbb{Q}_\ell$ instead of $\mathbb{Q}$ in characteristic $p$). Therefore $f_*$ is bijective. But then $f$ cannot contract any positive dimensional subvariety $Z$, because we would have $f_*[Z]=0$.
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4$\begingroup$ Since you're using $f_*$, you need to assume that $X$ is smooth, or perhaps use intersection homology. $\endgroup$ Aug 7, 2014 at 10:53
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2$\begingroup$ One can define $f_*$ on $N_1(X)$, the (real) vector space of 1-cycles up to numerical equivalence, which is finite-dimensional. Surjectivity then implies injectivity. $\endgroup$ Aug 7, 2014 at 11:58
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$\begingroup$ DCV: but this begs the question what is $f^*$ on $N_1$? I think you need some form of duality here. $\endgroup$ Aug 7, 2014 at 13:27
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2$\begingroup$ @DonuArapura: I think one can prove that $f_*$ is surjective by elementary means; all you need to do is, given a curve in $X$, find a 1-cycle mapping finitely to that curve via $f$. In particular there's no need to talk about $f^*$. Am I being too simple-minded? $\endgroup$– user5117Aug 7, 2014 at 14:00
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