I am reading A. Yekutieli's [original article](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=Yekutieli,%20A%2a&s5=dualizing%20complexes&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=9&mx-pid=1195406) on dualizing complexes for noncommutative algebras and I found a problem I cannot solve.


First, some background. We start with a field $k$ and a noetherian $\mathbb N$-graded algebra $A$ and assume that $A\_0 = k$. All objects are graded, and we write $\mathcal A$ for the category of graded $A$-*bimodules*. A dualizing complex is an object $R^\bullet$ in the derived category $\mathcal D(\mathcal A)$ satisfying the following:



> 
> 1. Its cohomology modules are finitely generated as left and right $A$-modules.
> 2. If we look at $R^\bullet$ as a complex of left $A$-modules, then it has finite injective dimension, and the same holds exchangng left for right.
> 3. The obvious morphisms $A \to \mathcal R Hom\_{A^\circ}(R^\bullet, R^\bullet)$ and $A^\circ \to \mathcal R Hom\_{A}(R^\bullet, R^\bullet)$ are isomorphisms in the category $\mathcal D (\mathcal A)$.
> 
> 
> 


These "obvious morphism" come from the following setup: By 2 we may replace $R^\bullet$ with a bounded complex $I^\bullet$ such that each $I^i$ is injective as left and right $A$-module (but not necessarily as bimodule). Thus for every complex of left $A$-modules $M^\bullet$ there is a map, defined at the level of chain complexes
\begin{align\*}
 M^\bullet \to Hom\_{A^\circ}(Hom\_A(M^\bullet, I^\bullet), I^\bullet)
\end{align\*}
given by simple evaluation. Passing to the derived category, the complex on the r.h.s. is isomorphic to $\mathcal R Hom\_{A^\circ}(\mathcal R Hom\_{A}(M^\bullet, R^\bullet), R^\bullet)$. Putting $M^\bullet = A$ gives us the morphism of point 3 of the above definition (if $M^\bullet$ is a complex of bimodules then the map respects this structure), and in the paper Yekutieli proves that this map is an isomorphism for $A$ if and only if it is an isomorphism whenever $M^\bullet$ is a bounded complex with finitely generated cohomology modules. So far, so good.


Dualizing complexes, when they exists, are not unique. In order to fix a special class of isomorphisms Yekutieli introduces local cohomology. Set $\Gamma\_m(-) = \varinjlim Hom\_A(A/A\_{\geq n},-)$, where $A\_{\geq n}$ denotes the ideal of elements of $A$ of degree at least $n$. This is a left exact functor, with a right derived functor $\mathcal R \Gamma\_m$ defined over left bounded complexes of left $A$-modules. Then he states that there is a map similar to that above,
\begin{align\*}
 \mathcal R \Gamma\_m(M^\bullet) \to \mathcal R Hom\_{A^\circ}(\mathcal R Hom\_{A}(M^\bullet, R^\bullet), \mathcal R \Gamma\_m(R^\bullet))
\end{align\*}
It is this map I am trying to make sense of. By looking at it simply at the level of complexes, we might feel tempted to replace $M^\bullet$ by an injective resolution $J^\bullet$, reuse our old resolution of $R^\bullet$ called $I^\bullet$, and thus obtain a map
\begin{align\*}
 \Gamma\_m(J^\bullet) \to Hom\_{A^\circ}(Hom\_{A}(J^\bullet, I^\bullet), \mathcal \Gamma\_m(I^\bullet))
\end{align\*}
given again by evaluation. This is indeed well-defined, but the problem is that passing to the derived category, it is not immediate, at least to me, that the complex on the r.h.s. is isomorphic to $\mathcal R Hom\_{A^\circ}(\mathcal R Hom\_{A}(M^\bullet, R^\bullet), \mathcal R \Gamma\_m(R^\bullet))$. In order for that to happen, $\Gamma\_m(I^\bullet)$ should consist of $A$-module which are injective as *right* $A$-modules, and I see no reason for that to be true (in the article it is proven that these modules are injective as *left* $A$-modules).


Any ideas on how to fix this? Have I made a mistake with the definition of this last morphism? 



Balanced dualizing complexes according to A. Yekutieli
I happened to be looking at this very issue a few days ago! Here is my explanation to myself... 


Indeed, in the notation above, there is a homomorphism of complexes of graded modules (not derived) 
$$ \zeta : \Gamma\_{\mathfrak{m}} (J) \to 
\mathrm{Hom}\_{A^{\mathrm{op}}} ( \mathrm{Hom}\_A ( J, I ), \Gamma\_{\mathfrak{m}} (I) ) . $$ 
And 
$$ \Gamma\_{\mathfrak{m}} (J) \cong \mathrm{R} \Gamma\_{\mathfrak{m}} (M) . $$
But now we can choose a quasi-isomorphism 
$ \Gamma\_{\mathfrak{m}} (I) \to K $
where $K$ is a complex of bimodules, injective on the right (and bounded below). We then get a morphism
$$ \eta : 
\mathrm{Hom}\_{A^{\mathrm{op}}} ( \mathrm{Hom}\_A ( J, I ), \Gamma\_{\mathfrak{m}} (I) ) \to
 \mathrm{Hom}\_{A^{\mathrm{op}}} ( \mathrm{Hom}\_A ( J, I ), K) \cong 
\mathrm{RHom}\_{A^{\mathrm{op}}} ( \mathrm{RHom}\_A ( M, R ), \mathrm{R}\Gamma\_{\mathfrak{m}} (R) ) . $$
The morphism we want is $\eta \circ \zeta$. It is functorial. 


In case $A$ is noetherian, we know that 
$ \Gamma\_{\mathfrak{m}} (I) $ is already injective on the right (as mentioned in the previous answer), and then $\eta$ is an isomorphism. 


I hope this makes sense! 


Added 27 Nov 2018: 


1. When $A$ is noetherian connected, and $R$ is a balanced DC, then $\mathrm{R} \Gamma\_{\mathfrak{m}}(R)$ is isomorphic to $A^{\*}$, and this is graded-injective on both sides.
2. In the book (in preparation) [**Derived Categories** (prepulication ver 3)](https://arxiv.org/abs/1610.09640) there is an improved treatment of local duality in the NC connected graded case -- see Thm 17.2.7.


