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In the answer to the question "Mathematically mature way to think about Mayer–Vietoris" given by Angelo, he claims that the right derived functors $R^k(j)$ of the left-exact embedding $$j:Sh~X\subseteq PreSh~X$$ are given for a sheaf $F$ by the presheaf $R^k(j)(F)$ sending an open set $V\subseteq X$ to $H^k(U,F)=H^k(U,F|_U)$. The latter should be sheaf cohomology, I think. How can I see that the right derived functors $R^k(j)$ are actually sheaf cohomology?

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This is basically an exercise in manipulating the definition of right derived functors via injective resolutions.

Write $\check{H}{}^*(U, -)$ for the right derived functors of $\Gamma (U, -) : \textbf{AbPsh}(X) \to \textbf{Ab}$, $\mathscr{H}^*(-)$ for the right derived functors of the inclusion $j_* : \textbf{AbSh}(X) \to \textbf{AbPsh}(X)$, and $H^*(U, -)$ for the right derived functors of $\Gamma (U, -) : \textbf{AbSh}(X) \to \textbf{Ab}$. We make the following observations:

  1. The functor $\check{H}{}^0(U, -)$ is exact.

  2. If $C^\bullet$ is an injective resolution of $A$ in $\textbf{Sh}(X)$, then $H^* (U, A) \cong H^*(\Gamma(U, C^\bullet))$ and $\mathscr{H}^* (A) \cong H^* (j_* C^\bullet)$.

  3. But $\check{H}{}^0(U, -)$ commutes with cohomology, so $$H^* (U, A) \cong H^*(\Gamma(U, C^\bullet)) \cong \check{H}{}^0(U, H^*(j_* C^\bullet)) \cong \check{H}{}^0(U, \mathscr{H}(A))$$ hence $\mathscr{H}(A)$ must be the presheaf $U \mapsto H^* (U, A)$.

(If you're wondering about the notation $\check{H}{}^* (U, -)$, this is because this is a special case of Čech cohomology with respect to the trivial open cover $\{ U \}$. More generally one may formulate Čech cohomology with respect to an open cover $\mathfrak{U}$ as the right derived functors of the functor $\textrm{Match}(\mathfrak{U}, -) : \textbf{AbPsh}(X) \to \textbf{Ab}$ that sends a presheaf $P$ to the set of matching families of sections of $P$ over $\mathfrak{U}$.)

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  • $\begingroup$ Thank you for the answer. I don't understand the second isomorphism $H^*(\Gamma(U,C^\bullet))\cong \check{H}^0(U,H^*(j_*(C^\bullet)))$. Moreover, isn't $\check{H}^0(U,-)=\Gamma(U,-):\mathbf{AbPsh(X)}\to \mathbf{Ab}$ as the zeroth derivative of a functor coincides with the functor? And isn't $\Gamma$ only left-exact? $\endgroup$ – Ronald Bernard Feb 18 '13 at 8:12
  • $\begingroup$ As I said, this is because $\check{H}{}^0(U, -)$ commutes with cohomology. Yes, $\check{H}{}^0(U, -) = \Gamma (U, -)$, but this $\Gamma$ is a functor on presheaves, not sheaves! (Recall, colimits of presheaves are computed differently. It turns out that taking sections of presheaves is exact.) $\endgroup$ – Zhen Lin Feb 18 '13 at 8:14
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    $\begingroup$ Yes, that's what I said. $\endgroup$ – Zhen Lin Feb 18 '13 at 8:38
  • $\begingroup$ If $\Gamma:\mathbf{AbShv}(X)\to \mathbf{Ab}$ and $\bar\Gamma:\mathbf{AbPsh}(X)\to \mathbf{Ab}$ and $\Gamma(U,-)=\bar\Gamma(U,j_*(-))$, then $H^*(\Gamma(U,C^\bullet))\cong H^*(\bar\Gamma(U,j_*(C^\bullet)))$ is the same as $\bar\Gamma(U,H^*j_*(C^\bullet))$ because $\bar\Gamma(U,-)$ commutes with cohomology, correct? $\endgroup$ – Ronald Bernard Feb 18 '13 at 8:40

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