# Are the right derived functors of the inclusion sheaf cohomology?

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?

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 $\mathfrak{U}$.)

• 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? – Ronald Bernard Feb 18 '13 at 8:12
• 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.) – Zhen Lin Feb 18 '13 at 8:14
• Yes, that's what I said. – Zhen Lin Feb 18 '13 at 8:38
• 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? – Ronald Bernard Feb 18 '13 at 8:40