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I am stuck with the following farce on derived Homs.

I have an abelian category $A$ and I showed that, given any two objects $X$ and $Y$ of $A$, the group of $1$fold extensions $\operatorname{Ext}^1_{A}(Y,X)$ is nonzero whenever both $X$ and $Y$ are nonzero. This yields that any injective (resp. projective) object in $A$ must be the $0$ object.

Yet, I am almost sure that the derived Hom exists in my case, mainly for the reason that follows. Assume $Y$ fixed. I proved the existence of a complex $C_Y(X)$ sitting in non-negative degrees, functorial in $X$, given with a natural transformation $$\varepsilon:\operatorname{Hom}_A(Y,-)[0]\longrightarrow C_Y(-)$$ inducing an isomorphism on $H^0$, and satisying the following property: for each short exact sequence $0\to X'\to X\to X''\to 0$ in $A$, there is a long exact-sequence: $$0\to H^0(C_Y X')\to H^0(C_Y X) \to H^0(C_Y X'') \to H^1(C_Y X')\to \cdots $$ naturally isomorphic to the canonical sequence $0\to \operatorname{Hom}_A(Y,X')\to \operatorname{Hom}_A(Y,X) \to \operatorname{Hom}_A(Y,X'') \to \operatorname{Ext}^1_A(Y,X)\to \cdots $, the first three vertical isomorphisms being induced by $\varepsilon(X')$, $\varepsilon(X)$ and $\varepsilon(X'')$.

My guess is that $\operatorname{RHom}_A(Y[0],X^{\bullet})$, if it exists, is quasi-isomorphic to $\operatorname{cone}(C_Y(X^{\bullet}))[-1]$. I imagine that I have to prove that $\operatorname{cone}(C_Y(-))[-1]$ satisfies the universal property of the right-derived functor of $\operatorname{Hom}_{C(A)}(Y[0],-)$.

Do you know how I can proceed? Many thanks!

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  • $\begingroup$ Some context would be very helpful. What is the definition of the category A in your case? $\endgroup$ May 23 at 19:31
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    $\begingroup$ Yes, it is possible to compute derived functors such as RHom without injectives or projectives. The theory of cotorsion pairs lets you construct derived categories (plus RHom) in great generality. For example, you can use flat modules. This was spelled out by Hovey in Cotorsion pairs, model category structures, and representation theory (e.g., see Theorem 2.5) and extended in great generality by Gillespie in many papers. For example his recent paper "K-flat complexes and derived categories" does it from K-flat complexes or indeed from any class of chain complexes as input. $\endgroup$ Jun 11 at 10:03

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Some construction of derived Hom complexes in an arbitrary $k$-linear Quillen exact category (for any commutative ring $k$) is worked out in the appendix to my paper "Artin-Tate motivic sheaves with finite coefficients over an algebraic variety", Proc. London Math. Soc. 111 (2015), #6, p.1402-1430, https://doi.org/10.1112/plms/pdv057 , https://arxiv.org/abs/1012.3735 . Other approaches may well exist in the literature, but I am not aware of them.

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    $\begingroup$ Dear @Leonid, thank you for your relevant answer! Seems indeed that I am in possession of the complex $C_{\mathcal{E}}$ you are describing there. Something that I do not quite get is how your construction recovers $RHom_{\mathcal{E}}(Y,-)$ whenever the latter exists, and whether it can be used to prove that $RHom_{\mathcal{E}}(Y,-)$ indeed exists. When I refer to $RHom_{\mathcal{E}}(Y,-)$, I mean the the unique functor satisfying the "universal property of the right derived functor of $Hom_{\mathcal{E}}(Y,-)$". Do you have some thoughts on that? $\endgroup$
    – Stabilo
    May 24 at 8:04
  • $\begingroup$ @Stabilo I am not sure that I understand the question. To my understanding, the universal property of the right derived functor is a property of the cohomology rather than of the complexes computing the cohomology. It is satisfied by the $Ext$ groups, not by any $RHom$ complexes. $\endgroup$ May 24 at 9:16
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    $\begingroup$ Let me precise: let $F:A\to B$ be a left-exact functor of abelian categories, and let $Q:K(A)\to D(A)$ be the localization functor (resp. $K(B)\to D(B)$). We say the pair $(RF,\varepsilon)$ - where $RF:D(A)\to D(B)$ is a functor and $\varepsilon:Q\circ F\to RF\circ Q$ a natural transformation - is a right derived functor of $F$ if, for each pair $(G,\eta)$ as above, there exists a unique natural transformation $\alpha:RF\to G$ such that $\eta=\alpha\circ \varepsilon$. Is your complex $C_{\mathcal{E}}(X,-)$ - or rather its shifted cone? - a right derived functor of $Hom_{\mathcal{E}}(X,-)$? $\endgroup$
    – Stabilo
    May 24 at 9:30
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    $\begingroup$ This way, the universal property of the right derived functor is still not a property of any complexes computing it, but rather of the derived category objects these complexes represent, i.e., of the quasi-isomorphism classes of these complexes. Defining and computing $RHom$ up to quasi-isomorphism should be much easier than constructing actual well-behaved complexes computing it. Concerning your question, I guess that the answer should be positive, but right now I cannot immediately prove it off the top of my head. $\endgroup$ May 24 at 10:56

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