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Let $Sym^m(X)$ be the $m$th symmetric product of a smooth projective variety $X$, $n=\dim(X)$, $Y_1$ an ample hypersurface of $X$, and $CH_0(X)_{hom}$ the Chow groups of $0$-cycles of degree $0$.

I read the following in the proof of a proposition in Voisin's book "Hodge Theory and Complex Algebraic Geometry II", page 286:

We thus conclude that a general fiber of $\sigma_m^{P_0}:Sym^m(X)\rightarrow CH_0(X)_{hom}$ intersect $Sym^m(Y_1)$ for sufficiently large $m$ and $n\geq 2$.

This means that \begin{equation*} \sigma^{P_0}_m:Sym^m(X)\rightarrow CH_0(X)_{hom} \end{equation*} and \begin{equation*} \sigma^{P_0}_m\arrowvert_{Sym^m(Y_1)}:Sym^m(Y_1)\rightarrow CH_0(X)_{hom} \end{equation*} have the same image for a sufficiently large $m$.

But I can not see how to get this implication, so I would be grateful if someone can give any advise about it. Thank you!

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  • $\begingroup$ I suppose you are asking for an explanation of "This means that..."? You must prove that a general element $\mathfrak{z}\in\operatorname{Sym}^m(X) $ is linearly equivalent to an element of $\operatorname{Sym}^m(Y_1) $. This is exactly saying that the fiber of $\sigma ^{P_0}_{m}$ containing $\mathfrak{z}$ contains an element of $\operatorname{Sym}^m(Y_1) $, i.e. intersects this subspace. $\endgroup$
    – abx
    Commented Apr 15, 2022 at 14:03
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    $\begingroup$ P.S: This is the second question you ask about Voisin's book, and both had a very simple answer. I think you should try to think more before posting such questions here. $\endgroup$
    – abx
    Commented Apr 15, 2022 at 14:10

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