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Let $X$ be a complex projective variety, therefore exist an embedding $i:X \to \Bbb{P}^n_\Bbb{C}$, therefore it will induce a functor $i_*$ from the coherent sheaf on $X$ to coherent sheaf on $\Bbb{P}^n_{\Bbb{C}}$. Since is right adjoint to the $i^*$. Therefore $i_*$ is left exact.

I was reading Algebraic geometry over the complex numbers by Arapura, when introduce the famous result that the cohomology group of coherent sheaves on projective variety are finite dimension, the book reduces it to the case that $X = \Bbb{P}^n_{\Bbb{C}}$ by $$H^{i}(X,\mathcal{E}) = H^{i}(\Bbb{P}^n,i_*\mathcal{E})$$

since $i_*$ is an exact functor.(where $i_*\mathcal{E}$ as direct image of coherent sheaf is again coherent.)

The question is I can't figure out why $i_*$ is right exact also? Does it depend on the projective condition?

yi li
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  • Oh I found it on stack project: https://stacks.math.columbia.edu/tag/04CJ . It depends on the closed immersion condition not projective condition. – yi li Mar 15 '23 at 04:32
  • this version is better: https://stacks.math.columbia.edu/tag/01QY – yi li Mar 15 '23 at 04:37
  • In general for topological space :Given an open inclusion $j:U\to X$ and $i:Z \to X$ with $Z = X\setminus U$ be the closed subset, we have $i_$ and $j_!$ being the exact functor, the key point is that $i_$ is left adjoint to $i^!$ and $j_!$ left adjoint to $j^{-1}$ – yi li Mar 15 '23 at 05:03

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