Construction for a push-forward of a module over a category

Question

This question is inspired by Example 2.3.13 in Weibel's Homological Algebra book. Perhaps it is too elementary for MO, but hopefully someone can give me a quick answer.

The example is the following: let $I$ be a small category and $A$ an abelian category and $A$ is complete. Let $M \in A^I$ be a functor $M: I \to A$ (morally a module over $I$, not sure about the terminology). For any $k \in I$, we have the $k^{th}$ coordinate $ev_k: A^I \to A$ given by $M \mapsto M(k)$. The construction in the example gives a right adjoint to $ev_k$, denoted as $k_*: A \to A^I$, namely for any $a \in A$, we define $$ k_*(a)(i) = \prod_{\hom_I(i,k)} a $$ This seems to me to be like a skyscraper sheaf over the point $k$.

My question is the following: Let $I, J$ be small categories, and $A$ a complete abelian category. Fix a functor $F: J \to I$, then we have the pull-back functor $F^*: A^I \to A^J$. Can we define the right adjoint $F_*$ of $F^*$? That is, for any $M \in A^I$, $N' \in A^J$, we have $$Hom_{A^J}(F^*(M), N') = Hom_{A^I}(M, F_*(N'))$$

Answer 1

In more generality, let $\mathbf{C}$, $\mathcal{I}$, and $\mathcal{I}^\prime$ be categories, with a functor $F:\mathcal{I}\to\mathcal{I}^\prime$. Composition induces a map $\overline{F}:\mathbf{C}^{\mathcal{I}^\prime}\to\mathbf{C}^\mathcal{I}$. If this map has a left adjoint, the left adjoint is known as a left Kan extension along $F$. The image of a functor $g:\mathcal{I}\to\mathbf{C}$ under this left adjoint is called the left Kan extension of $g$ along $F$. Likewise for right adjoints. The left (resp. right) Kan extension along a functor $F:\mathcal{I}\to\mathcal{I}^\prime$ if $\mathcal{I}$ is small and $\mathbf{C}$ admits all small colimits (resp. limits). You can learn more in a category theory textbook.