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Let $X$ be a smooth complex projective variety, and $Y\hookrightarrow X$ be a smooth projective subvariety. Let $\pi:\tilde{X}\rightarrow X$ be the blow-up along $Y$, and let $j:E\hookrightarrow \tilde{X}$ be the exceptional divisor.

Let $Z$ be a $k$-dimensional subvariety of $E$ such that $dim\,\pi(Z)=dim\,Z$. Now, $[Z]\in CH_k(E)$ is a $k$-cycle, and let $Z':=j_*([Z])$.

Question: What is the relation between $\pi^*\pi_*(Z')$ and $Z'$ ? Are they equal?

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    $\begingroup$ No : Try $Z=E$. $\endgroup$
    – abx
    Commented Jul 17, 2019 at 19:53
  • $\begingroup$ @abx: Thanks! I forgot to exclude the case when $\pi_*$ might become zero. I have edited the question. $\endgroup$
    – Hajime_Saito
    Commented Jul 17, 2019 at 20:11

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Say $Y$ is $\mathbb P^1$ and has codimension $2$, so $E$ is a $\mathbb P^1$-bundle on $\mathbb P^1$. Then this $\mathbb P^1$-bundle has many sections, which are not equivalent in the Chow group, but only equivalence class contains pullback of $Y$, which is the pullback of the pushforward of any section.

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  • $\begingroup$ Sir, I apologize that my knowledge of Chow groups is rather limited; I couldn't understand the part "...but only equivalence class contains pullback of Y, which is the pullback of the pushforward of any section". Can you please explain this part? $\endgroup$
    – Hajime_Saito
    Commented Jul 17, 2019 at 20:55
  • $\begingroup$ @yojusmath If $Z$ is a section, then $\pi_* Z = Y$, so $\pi^* \pi_* Z = \pi^* Y$. This just comes from the definition of $\pi_*$. $\endgroup$
    – Will Sawin
    Commented Jul 17, 2019 at 21:01

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