# Is $\mathcal{Ext}^i(F,G)$ the sheafification of $Ext^i(F,G)$?

Let $F,G$ be quasicoherent sheaves of modules on a scheme $X$, then is the sheaf $\mathcal{Ext}^i(F,G)$ equal to the sheafification of the presheaf $$E^i(F,G) : U\mapsto Ext^i_U(F|_U,G|_U)?$$

It seems to me that modulo some technical points, this is probably true.

Indeed, let $G\rightarrow I^\bullet$ be a injective resolution for $G$. By Lemma III.6.1 in Hartshorne's Algebraic Geometry, we have that for any open $U\subset X$, $I^\bullet|_U$ is an injective resolution for $G|_U$. (Actually this result seems somewhat suspicious to me, since I have heard multiple times that "localization does not always preserve injectives").

But, assuming Hartshorne's lemma is correct, the cohomology of the complex $Hom_U(F|_U,I^\bullet|_U)$ gives precisely the groups $Ext^i_U(F|_U,G|_U)$.

On the other hand for any inclusion of opens $U\subset V$, localizing yields a morphism of complexes $$Hom_V(F|_V,I^\bullet|_V)\rightarrow Hom_U(F|_U,I^\bullet|_U)$$ and hence a map on cohomology, which is clearly functorial w.r.t. further localization. Thus, for any inclusion of opens $U\subset V$, this gives restriction maps

$$Ext^i_V(F|_V,G|_V)\rightarrow Ext^i_U(F|_U,G|_U)$$ which shows that the rule $U\mapsto Ext^i_U(F|_U,G|_U)$ defines a presheaf.

Now, we want to define a map of presheaves $E^i(F,G)\rightarrow \mathcal{Ext}^i(F,G)$. To do this, we note that we have an equality of complexes $$Hom_U(F|_U,I^\bullet|_U) = \Gamma(U,\mathcal{Hom}_U(F|_U,I^\bullet|_U))$$ If $\partial$ denotes the differential of the complex $\mathcal{Hom}_U(F|_U,I^\bullet|_U)$, and $d$ denotes the differential of the complex $Hom_U(|F_U,I^\bullet|_U)$, then we get a map $$\ker d^i\stackrel{=}{\longrightarrow}\Gamma(\ker\partial^i)\rightarrow\Gamma(\ker\partial^i/\text{im }\partial^{i-1})$$ which factors through $\text{im }d^{i-1}$, and hence for every open $U$, we get compatible maps $$Ext^i_U(F|_U,G|_U)\rightarrow\Gamma(U,\mathcal{Ext}^i(F,G))$$ whence a morphism of presheaves $$\varphi : E^i(F,G)\rightarrow\mathcal{Ext}^i(F,G)$$ which seems to induce isomorphisms on open affines (is there anything to check here? must we impose any finiteness conditions like $X$ noetherian or $F$ coherent?).

If the above is correct, then $\varphi$ should induces an isomorphism on stalks, and hence identifies $\mathcal{Ext}^i(F,G)$ with the sheafification of $E^i(F,G)$.

Can someone check this proof and/or confirm the result? References would be appreciated as well.

• Hi Amy, this idea may be complete garbage, but there is a very similar result for higher direct images (Hartshorne III 8.1). I can't see any reason why that method of proof can't be transferred here - though I would love to hear what you think! – Kenny Wong Mar 29 '17 at 1:43
• By the way, this might be helpful: stacks.math.columbia.edu/tag/0BQP – Kenny Wong Mar 29 '17 at 10:19
• @KennyWong I'm a bit confused about what they (in your stacksproject link) mean by $Ext^n_{D(\mathcal{O}_U)}(K|_U,L|_U)$. How is that different from $Ext^n_{\mathcal{O}_U}(K|_U,L|_U)$? – user355183 Mar 29 '17 at 18:04
• I believe $D(\mathcal O_X)$ in the stacksproject link means the derived category of coherent sheaves on $X$. So the sheaf $K$ is identified with the complex $\dots \to 0 \to K \to 0 \to \dots$, which is an object in $D(\mathcal O_X)$. Ultimately, their $Ext^n_{D(\mathcal O_U)} (K|_U, L|_U)$ is equal to your $Ext^n_{\mathcal O_U} (K|_U, L|_U)$ - unless I'm terribly mistaken. – Kenny Wong Mar 30 '17 at 1:04
• As far as I know, they are only related via a spectral sequence. See proposition 14 of therisingsea.org/notes/SpectralSequences.pdf – Justine Apr 2 '17 at 22:25