Let $X$ be a smooth curve over a field. Let $Y$ be the triple product $X \times X \times X$. Let $\gamma$ be a homologically trivial codimension $2$ cycle.


In the text [Zhang, p. 76] that I am currently reading it is concluded that $\pi\_{i,\*}(\gamma)$ is trivial (rationally). I do not see why this is true, and could not find a proof elsewhere.


In [Zhang] there is more context, but I think that I stated the relevant input for the claim.


Reference
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[Zhang] Shou-Wu Zhang. *“Gross–Schoen Cycles and Dualising Sheaves”*. arXiv: <http://arxiv.org/abs/0812.0371> .




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*(I should add a note that I am currently not in the situation to search through a library. So maybe this can be found in any book on algebraic cycles; if so, please give the reference.)*



Why does a homologically trivial cycle have trivial projections?
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.


