Disclaimer: This was first asked [here on math.stackexchange](https://math.stackexchange.com/questions/2207822/is-mathcalextif-g-the-sheafification-of-extif-g) with no answers.


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 (maybe 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 kills $\text{im }d^{i-1}$, 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. On the one hand it seems like a nice statement to have in mind, and yet on the other hand if it were true, I would have expected to have seen/heard it somewhere.



Is $\mathcal{Ext}^i(F,G)$ the sheafification of $Ext^i(F,G)$?
This is proven in the [Stacks Project](http://stacks.math.columbia.edu/tag/0BQP) as pointed out in the [comments on math.stackexchange](https://math.stackexchange.com/questions/2207822/is-mathcalextif-g-the-sheafification-of-extif-g#comment4543299_2207822) by essentially the same proof you suggested.


