I am trying to understand the bifunctor $R\operatorname{Hom} : D(\mathcal{A}) ^{op} \times D(\mathcal{A}) \to D(\operatorname{Ab})$ (I am also interested in the total right derived functor of the enriched hom functor if $\mathcal{A}$ is a closed category, but I assume it is done in an analogous way). I think my confusion is with total derived functors in general.

In Charles Weibel's "An Introduction to Homological Algebra" there are only 2 pages on $R \operatorname{Hom}$ and they are very confusing. It seems like he uses the functor $\operatorname{Hom}^\bullet$, which is defined by

$$\operatorname{Hom}^\bullet (A, B) = \operatorname{Tot}(\{\operatorname{Hom}(A^p, B^{-q})\})$$

so that $R\operatorname{Hom} (A,B)$ is $\operatorname{Hom}^\bullet (A,B)$ whenever $A$ is a complex of projectives or $B$ is a complex of injectives.

However, wouldn't it be more accurate to name this $R\operatorname{Hom}^\bullet$? It does mention in a later exercise that $R \operatorname{Hom}(-,B)$ is equivalent to the total derived functor of $\operatorname{Hom}(-,B)$ if $B$ has finite injective dimesnion. However, I am not sure how to make any sense of this, since the book only defines total derived functors for morphisms of triangulated categories, where the target is the homotopy category of some abelian category, and I am not sure how to turn $\operatorname{Hom}(-,B)$ into such a morphism, since it maps to abelian groups, not complexes.

On the other hand, Gelfand and Manin's "Methods of Homological Algebra" only defined derived functors $RF: D(\mathcal{A}) \to D(\mathcal{B})$ for $F: \mathcal{A} \to \mathcal{B}$ a left exact functor that is exact when restricted to complexes of injectives. Then, they take $R \operatorname{Hom}(A,-)$ only when $A$ is in $\mathcal{A}$, rather than a complex. That is, if $B$ is made up of injectives, $R \operatorname{Hom}(A,B)$ is the complex obtained by applying $\operatorname{Hom}(A,-)$ at every degree. (You can also take the dual definition with projectives)

What is the "correct" way of defining $R \operatorname{Hom}$? If anyone can also refer me to resources on this it would be hugely appreciated (I have looked at all of the usual recommended books for homological algebra and none of them seem to go into this)

  • $\begingroup$ $RHom$ is only defined up to quasi-isomorphism, are you asking for a reference to the fact that these definitions coincide up to q-iso? $\endgroup$ Jan 3, 2021 at 14:22

1 Answer 1


The bifunctor $$R\operatorname{Hom}^\bullet \colon D(\mathcal{A}) ^{op} \times D(\mathcal{A}) \to D(\operatorname{Ab}) $$ is understood —whenever $\mathcal{A}$ has enough injective objects (e.g. $\mathcal{A}$ is a Grothendieck category)— as the right derived functor of the functor $$\operatorname{Hom}^\bullet \colon K(\mathcal{A}) \to K(\operatorname{Ab}) $$ with the first variable fixed: accordingly one uses the following to compute it: $$ R\operatorname{Hom}^\bullet(A, B) = \operatorname{Hom}^\bullet(A, I_B) $$ where $I_B$ denotes a k-injective resolution of $B$. It makes sense for $A \in D(\mathcal{A})$ and the fact that preserves products in the first variable follows from the universal property. See this answer.

Another issue is what happens if moreover $\mathcal{A}$ has enough projective objects. In this case there is another functor defined as

$$ R'\operatorname{Hom}^\bullet(A, B) = \operatorname{Hom}^\bullet(P_A, B) $$

Again, by the universal property it is easy to prove that $R\operatorname{Hom}^\bullet$ is balanced, i.e. there is a canonical isomorphism: $$ R\operatorname{Hom}^\bullet(A, B) \cong R'\operatorname{Hom}^\bullet(A, B) $$

As for notation, one writes $R\operatorname{Hom}^\bullet(-,-)$ to emphasize we are dealing with the functor hom that takes value in complexes, unfortunately sometimes the authors are not consistent with this convention.


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