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I wonder whether following statements holds

If A is an abelian category(or quasi abelian category) having enough projectives, then category of pointed diagram(which means diagram has final object,or for simplicity, one assume the diagram is finite)(A_D=(D--->A))has enough projectives.

I want to construct a pair of adjoint functor between this two category. Then use left adjoint of exact functor maps projectives to projectives

Other methods to prove this statement is welcomed

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3 Answers 3

up vote 2 down vote accepted

If D is small and A has enough projectives and has infinite sums then $A^D$ has enough projectives. For the proof, see Weibel, "An introduction to homological algebra", 2.3.13 on p.43. It contains the adjoint you are apparently looking for.

The proof is a version of Godement's argument that the category of sheaves of abelian groups has enough injectives.

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There are several pairs of adjoint functors of the kind you desire but it isn't clear to me if any (or all) of them will give you enough projectives in $A^D$.

For example for each $d$ in $D$ $\iota_d$ which sends an object of $A$ to the diagram that is $A$ at $d$ and zero elsewhere and does the obvious thing on morphisms is left adjoint to the functor $\pi_d$ that sends a diagram to its value at $d$.

Also if $D$ is pointed and $A$ has $D$-colimits then $\mathrm{colim}\colon A^D\rightarrow A$ is left-adjoint to the constant functor that sends $X$ in $A$ to the diagram that is $X$ everywhere and all morphisms are $\mathrm{id}_X$ (see Wikipedia).

These left adjoints will all map projectives to projectives.

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Thank you, I will check the details –  Shizhuo Zhang Nov 25 '09 at 15:02
    
The statement beginning "For example..." isn't quite right, ι_d has to send an object e to something like Hom(d, e)·A. –  Reid Barton Nov 25 '09 at 16:58
    
You're right. I realised this when writing my answer and deleted it. Then I put it back again having forgotten that's why I deleted it. The colimt example is correct though. Of course it is all rather irrelevant for this question now as the correct answer is above. –  Simon Wadsley Nov 26 '09 at 8:42

As Valery Alexeez wrote From Weibel, "An introduction to homological algebra", 2.3.13 on p.43 follow that $\mathcal{A}^I$ has enought projectives. But isnt clear if for a projective $P\in \mathcal{A}^I$ each $P(i)\in \mathcal{A}$ is projective as we need further.

(this is true if for any $i\in I$ the right Kan extention of the $i$-valutation $v(i): \mathcal{A}^I\to \mathcal{A}$ is exact, I dont know if this follow from the "$I$ is filtrant" hypothesis).

Alternatively:

From the book "Theory of Categories (BArry Mitchell) cor.7.6 p. 138, let $T_i: \mathcal{A}^I\to \mathcal{A}$ ($i\in I$) the $i$-valuation, and $S_i$ its left adjuction (the left Kan extention), now for a projective $P\in \mathcal{A}$ the object $S_i(P)(j),\ j\in I$ is a sum of copies of $P$ (see how the left KAn extention is maked, for example in the Weibel reference above) then is projective. THe above corollary asserts that projectives of $\mathcal{A}^I$ are objects of the form $\bigoplus_{i\in I}S_i(P_i)$ (where $P_i\in \mathcal{A}$ is a prjective) and all its retracts.

This ensure the existence a projective resolution of a diagram $(X_i)_i\in \mathcal{A}^I$ with projective arguments.

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