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Embed $X$ in $\mathbb{P}^N$. If $N>\dim X$, find a point $p$ not on $X$ and project from that point. You will get a finite morphism from $X$ onto $X'\subseteq \mathbb{P}^{N-1}$, and so on. Am I missing something?
Yes, this is true, and I think general linear projection is the usual argument. It works over algebraically closed fields. For CM varieties, the morphism is flat and you can obtain a nice proof of Serre duality using duality for finite flat morphisms...
If we replace $t^2$ by $t$, the map $\phi$ becomes the $9$-uple embedding of $\mathbb{P}^1$. In particular, it is a closed immersion and the image is smooth.
I support this question being posted on MO. I think some of the users here might have a better feeling of what a mathematician needs to learn to have good intuitions in string theory-related mathematics. Also, I wanted to ask such a question myself ;)
Then I guess this is just a descent question for $\bar\mathbb{Q}_l$-sheaves and morphisms between them, and the derived category doesn't play a role here. I think this should be true and that you should just push $K$ forward to $X/A$ and take invariants.
You can still hope that the following holds: suppose $f:X\to Y$ and $\mathcal{F}$ are like in your first sentence. Then if $R^i f_*(\mathcal{F})$ is locally free on $Y$, then $R^i f_*(\mathcal{F})$ commutes with base change. This I think is true and follows from the discussion of cohomology and base change in Mumford's "Abelian varieties".
What exactly do you mean by "$K$ is equivariant with respect to $A$" if $K$ is just an object in the derived category? Is $a^* K$ required to be isomorphic, or just quasi-isomorphic to $K$?
Regarding your comment above: assume $X$ is smooth. Then $I_Z$ a line bundle if and only if $Z$ is a divisor (consists only of curves), which I believe is both open and closed...
