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Suppose we have an Abelian category $\mathfrak A$ and a ring $R$. From this data we can form a new Abelian category $\mathfrak A[R]$ whose objects are objects $A\in\mathfrak A$ together with a ring homomorphism $R\to\mathfrak A(A,A)$ and whose morphisms are morphisms in $\mathfrak A$ commuting with the $R$-actions.

Let's assume that we understand the category $\mathfrak A$ and the ring $R$ well in the sense that we know the projective objects in $\mathfrak A$ and in the category of $R$-modules. Can we use this to characterise projective objects in $\mathfrak A[R]$ in a nice way?

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Do we understand $\mathfrak{A}$ well enough to be able to say that the forgetful functor $\mathfrak{A}[R] \to \mathfrak{A}$ has a left adjoint $L$? This left adjoint will preserve projectives. In particular, if $\mathfrak{A}$ has enough projectives then the same will be true for $\mathfrak{A}[R]$. If the forgetful functor has in addition an exact right adjoint then every projective in $\mathfrak{A}[R]$ will be a direct summand of an object of the form $L(P)$ with $P$ projective in $\mathfrak{A}$. I don't know if that helps. – t.b. Mar 19 '11 at 21:53
@Theo: for $\mathfrak A=\mathrm{Abelian\ groups}$ this would be tensoring with $R$, right? In the case I am interested in, $\mathfrak A=\mathfrak{Coh(T)}$, where $\mathfrak T$ is a triangulated category and $\mathfrak{Coh(T)}$ is the Abelian category of finitely represented additive cofunctors functors $\mathfrak T\to\mathrm{Abelian\ groups}$. The projective objects in there are the (retracts of) represetable functors. I don't see how to write down a left adjoint to the forgetful functor here. – Rasmus Mar 19 '11 at 22:54
Yes, of course. And I suspect $\mathfrak{T}$ is $KK$ (given that you've asked about $C^{*}$-stuff a few times)? Oh, my. Do you really have to work with that beast $\mathfrak{Coh(T)}$? In my experience it is not very maniable. What's $R$ by the way? – t.b. Mar 19 '11 at 23:11
@Theo: You're guessing correctly.^^ One example would be $\mathfrak T=KK$ and $R=\mathrm{Rep}(G)$ for a compact group $G$. Meyer and Nest show that you can make sense of the bar resolution in $\mathfrak{Coh}(KK)[\mathrm{Rep}\ G]$ which I find pretty awesome. I'm not sure if I have to work with that beast -- I just made its acquaintance... What would it mean to understand $\mathfrak T[R]$? – Rasmus Mar 19 '11 at 23:34
No surprise (the former, not the latter). But then you should have a candidate for the left adjoint! A bar resolution is usually of the form $\cdots \to LFLF X \to LF X \to X$ where $F$ is the forgetful functor. $L$ should be some sort of induction in any case. – t.b. Mar 19 '11 at 23:48

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