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A functor $F:T\to R$ between triangulated categories is dense if every object of $R$ is isomorphic to a direct summand in the image of $F$.

Let $R=T=D^b(\text{coh }X)$ for a variety $X$ and consider the functor $-\otimes \mathcal{V}$, $\mathcal{V}$ a vector bundle.

I do not understand the following claim: "$-\otimes\mathcal{V}$ is a is a dense functor, as any object $P\in D^b(\text{coh }X)$ is a summand of $(P\otimes V^\vee)\otimes V$."

Can anyone help?

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up vote 2 down vote accepted

What part of the claim you don't understand?

For any vector bundle $V$ the bundle $V\otimes V^\vee$ contains trivial 1-dimensional vector bundle (spanned by the section "Id"$\in V\otimes V^\vee$; the map in the opposite direction is the evaluation map). So any object $P\in D^b$ is a summand of the image of the object $P\otimes V^\vee$.

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Thanks Grigory. Are you saying $V\otimes V^\vee\to \mathcal{O}_X$ splits? I was not aware of that. – Carsten Jan 25 '12 at 14:32
Yes, exactly (it splits because this map has section coming from $V\otimes V^\vee\cong hom(V,V)\ni Id$). – Grigory M Jan 25 '12 at 19:12

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