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Let $\mathcal{I}$ be a small category and $\mathcal{A}$ an abelian category. If $\mathcal{A}$ is complete (that is, the product of any set of objects exists) and has enough injectives, how can I prove that the functor category $\mathcal{A}^{\mathcal{I}}$ has enough injectives?

I know that if $R$ is a right adjoint functor to an exact functor $L$ then $R$ preserves injective objects. This may be used to solve the problem; there is a functor $R:\mathcal{A}\rightarrow\mathcal{A}^{\mathcal{I}}$ satisfying the proposition above? If yes, how can I prove it?

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If $F(i)\to Q_i$ are inclusive R into injectives for every $i\in I$, then a functor $F:I\to A$ with the values $F(i)$ includes into the product $\prod_i i_*Q_i$. The functor $i_*Q_i$ has the universal property that functors into it are in natural bijection with maps into $Q_i$ in $A$. They're called the right Kan extensions of $Q_i$ along the functors $i:*\to Q$, where $*$ is the one-point category. The explicit formula is $i_*Q_i(j)=\prod_{f:j\to i} Q_i$, and if you don't want to deal with the abstraction of Kan extensions, you can probably prove the claim directly from that formula.

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  • $\begingroup$ I don't know Kan extensions and it seems very abstract to me for now. I edited my question; maybe you can help. $\endgroup$ – Rafael Holanda Oct 6 '17 at 7:10
  • $\begingroup$ The right Kan extensions I mentioned are right adjoint to the evaluation functions $A^I\to A$. $\endgroup$ – Kevin Carlson Oct 7 '17 at 19:59

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