1
$\begingroup$

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 $$Hom_{A^J}(F^*(M), N') = Hom_{A^I}(M, F_*(N'))$$

$\endgroup$
2
  • 5
    $\begingroup$ Yes, and you don't need the abelian category assumption; completeness is enough. Look up right Kan extension in Mac Lane's Categories for the Working Mathematician. $\endgroup$
    – Todd Trimble
    Jul 30 '16 at 17:45
  • 2
    $\begingroup$ There is also a left adjoint, namely left Kan extension. $\endgroup$ Jul 30 '16 at 18:32
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.