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Let $R$ be a ring, and consider the hom functor $\hom\colon Mod(R)^\text{op}\times Mod(R)\to Mod(R)$; the end of $\hom$ is well-known to be the set of endomorphisms (endonatural transformations) of the identity functor $1\colon Mod(R)\to Mod(R)$, i.e. the kernel $$ \int_M \hom(M,M)\cong \ker\left(\prod_{M\in Mod(R)} \hom(M,M)\xrightarrow{(-)\circ \phi - \phi\circ(-)} \prod_{\phi\colon M\to N}\hom(M,N)\right) $$ Now, what should the end of the derived functors $Ext^n\colon Mod(R)^\text{op}\times Mod(R)\to Mod(R)$ be?

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    $\begingroup$ Please expand, if necessary in an answer, or follow me in chat :) no doubt that one could try to make precise that intuition, but bad things happen since resolutions are not canonical. $\endgroup$ – Fosco Apr 26 '15 at 14:08
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    $\begingroup$ yeah, I don't really know what I'm talking about. $\endgroup$ – bananastack Apr 26 '15 at 14:46
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    $\begingroup$ Let us denotes $D$ the category finitely projective $R$-modules. May be it will be helpful to notice that there is a quasi-isomorphism of chain complexes between $\mathbf{R}Hom(D,D)(id,id)$ and $HH(D)$ where $\mathbf{R}Hom(-,-)$ is the the derived internal hom in the category of DG-categories and $HH(D)\cong HH(R)$ is the Hochschild cochain complex associated to $D$ viewed as dg category in an obvious way. In my opinion, that should be the correct formulation in the derived world. $\endgroup$ – Ilias A. Apr 26 '15 at 15:11
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    $\begingroup$ Another way of saying it: Write $I: Mod(R) \to D(R)$ for the embedding into the derived category. Then the end of $Ext^n$ should be the group of natural transformations $Nat(I,I[n])$. $\endgroup$ – Phil Tosteson Apr 27 '15 at 20:39
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    $\begingroup$ Maybe you want $R$ to be commutative? Otherwise Hom won't be contained in $R$-modules. Or $R$ is any ring, and consider functors valued in abelian groups? $\endgroup$ – HeinrichD Nov 12 '16 at 18:40

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