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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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  • $\begingroup$ Suppose you have such a functor. How are you going to use it to solve the problem? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 6, 2017 at 20:14
  • $\begingroup$ @მამუკაჯიბლაძე it depends of the functors, I think. Maybe something like coordinates and pick up products... $\endgroup$
    – Rafael
    Commented Oct 6, 2017 at 20:24
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    $\begingroup$ Right Kan extension will do it! Instead of just one R, you have one for each object of I: the functor R_i that's right adjoint to the "evaluation at i" functor $\endgroup$
    – John Wiltshire-Gordon
    Commented Oct 7, 2017 at 2:41

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