# How to construct pair of adjoint functors from category A to category A_D(category of diagrams)

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

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.

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.

• 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.