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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.2011-03-19
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    @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.2011-03-19
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    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?2011-03-19
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    @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 $\mathfrak T[R]$?2011-03-19
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    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.2011-03-19
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    @Theo: Your previous comment is very helpful (as is your first comment)! Could you please provide me with a reference for this conceptional view on the bar resolution?2011-03-20
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    Chapters 8.6/8.7 in Weibel (monads/triples and simplicial methods are the key words here), see also Beck's thesis: http://www.tac.mta.ca/tac/reprints/articles/2/tr2abs.html. As for the first comment this is just playing around with the fact that left adjoints to exact functors preserve projectives.2011-03-20
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    Just a few things on $\mathfrak{Coh(T)}$ (I assume idempotents split throughout, but as you have countable coproducts in $KK$ this won't bother you): The embedding $\mathfrak{T\to Coh(T)}$ is the universal homological functor on $\mathfrak{T}$. It is Frobenius abelian: projective=injective and enough of them. You have adjoints on the level of triangulated cats iff you have them on the level of $\mathfrak{Coh}$. Finally, you can find four equivalent descriptions in Ch 5 of Neeman's book (don't miss App. C to see how bad $\mathfrak{Coh}$ can be - very far from well-powered).2011-03-21
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    available at hopf.math.purdue.edu/Neeman/triangulatedcats.pdf2011-05-05

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