Skip to main content
added 540 characters in body
Source Link
Angelo
  • 27k
  • 6
  • 92
  • 112

If $X$ and $Y$ are smooth complete varieties and $f\colon X \to Y$ is a morphism, there is a pushforward in cohomology, which is Poincaré dual to the pullback. This pushforward is compatible with the pushforward on cycles; that is, the cohomology class of the pushforward of a cycle is the pushforward of the cohomology class of the cycle. Hence, the pushforward of a homologically trivial cycle is homologically trivial.

[Edit:] the poster wants a proof that $\pi_{i*}\gamma$ is trivial as in the Chow ring. If $\pi_i$ is the projection onto the $i^{\rm th}$ factor, then $\pi_{i*}\gamma$ has codimension~0, and the part of the Chow ring in degree 0 is $\mathbb Z$, so any class of cycles of degree 0 that is homologically equivalent to 0 is in fact 0. The alternative is that $\pi_i$ is a projection onto a product of the other two factors, but then the statement would be false, that is, $\pi_{i*}\gamma$ could very well be not rationally equivalent to 0.

If $X$ and $Y$ are smooth complete varieties and $f\colon X \to Y$ is a morphism, there is a pushforward in cohomology, which is Poincaré dual to the pullback. This pushforward is compatible with the pushforward on cycles; that is, the cohomology class of the pushforward of a cycle is the pushforward of the cohomology class of the cycle. Hence, the pushforward of a homologically trivial cycle is homologically trivial.

If $X$ and $Y$ are smooth complete varieties and $f\colon X \to Y$ is a morphism, there is a pushforward in cohomology, which is Poincaré dual to the pullback. This pushforward is compatible with the pushforward on cycles; that is, the cohomology class of the pushforward of a cycle is the pushforward of the cohomology class of the cycle. Hence, the pushforward of a homologically trivial cycle is homologically trivial.

[Edit:] the poster wants a proof that $\pi_{i*}\gamma$ is trivial as in the Chow ring. If $\pi_i$ is the projection onto the $i^{\rm th}$ factor, then $\pi_{i*}\gamma$ has codimension~0, and the part of the Chow ring in degree 0 is $\mathbb Z$, so any class of cycles of degree 0 that is homologically equivalent to 0 is in fact 0. The alternative is that $\pi_i$ is a projection onto a product of the other two factors, but then the statement would be false, that is, $\pi_{i*}\gamma$ could very well be not rationally equivalent to 0.

Source Link
Angelo
  • 27k
  • 6
  • 92
  • 112

If $X$ and $Y$ are smooth complete varieties and $f\colon X \to Y$ is a morphism, there is a pushforward in cohomology, which is Poincaré dual to the pullback. This pushforward is compatible with the pushforward on cycles; that is, the cohomology class of the pushforward of a cycle is the pushforward of the cohomology class of the cycle. Hence, the pushforward of a homologically trivial cycle is homologically trivial.