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The following object has turned up in my research recently, and it would be surprising (to put it mildly) if no one else has seen, used or studied it before. I am hoping for a name and some references with basic properties, if these already exist.

Setup

Start with a full subcategory $A$ of a category $B$ and a functor $F:A \to \text{Vect}$. By full subcategory category I mean that every $B$-morphism between two objects of $A$ is also an $A$-morphism, and by $\text{Vect}$ I mean the category of vector spaces over some fixed field with linear maps as morphisms. In general, one could replace the target of $F$ by anything that has limits and colimits. Here is the picture: $$ \begin{array}{ccc} A & \stackrel{F}{\longrightarrow} & \text{Vect} \\\\ \downarrow & ~ & \\\\ B & ~ & ~ \end{array} $$

Two Extensions

Now, the left Kan extension $LF:B \to \text{Vect}$ of $F$ along the inclusion of $A$ into $B$ may be defined at any object $b \in B$ as the colimit of the diagram $F\downarrow b$. This is precisely $F$-image of all morphisms in $B$ from objects of $A$ with target $b$.

On the other hand, one has the right Kan extension $RF:B \to \text{Vect}$ defined dually. In this case, we use the limit of the diagram $b \downarrow F$, which consists of $F$-images of morphisms in $B$ from $b$ to objects of $A$.

Since $A$ is full in $B$, it is well-known that both $RF$ and $LF$ will make the above diagram commute on the nose rather than requiring any non-trivial natural transformations, etc.

The Object

Clearly -- using the fact that $A$ is full -- every object $a$ in the diagram $F \downarrow b$ has arrows to every object $a'$ in $b \downarrow F$ arising from the $F$-image of the compositions $a \to b \to a'$ in $B$. By universal properties of limits and colimits, this induces a unique linear map $LF(b) \to RF(b)$ for each $b \in B$ and in fact produces a natural transformation $\eta:LF \Rightarrow RF$ between Kan Extensions.

What is $\eta$ called, and are there any references to it as well as its properties?

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  • $\begingroup$ In your first and second paragraph on "Two Extensions" $LK$ and $RK$ should probably be $LF$ and $RF$ respectively. Moreover, I do not see where you need the assumption that the category is thin (BTW, I do prefer term "degenerated"), a you can construct you $\eta$ by the following natural isomorphism (applied in the reversed order to the identity on $F$; $J\colon \mathcal{A} \rightarrow \mathcal{B}$ stays here for the inclusion): $$\hom(\int^A(\hom(J(A),-)\times F(A), \int_B F(B)^{\hom(-, J(B))})$$ by the definition of the object of natural trnasformations: $\endgroup$
    – Michal R. Przybylek
    Commented Apr 22, 2013 at 11:08
  • $\begingroup$ $$\int_C \hom(\int^A(\hom(J(A),C)\times F(A), \int_B F(B)^{\hom(C, J(B))})$$ the preservation of ends and coend: $$\int_{A,B,C} \hom(\hom(J(A),C)\times F(A), F(B)^{\hom(C, J(B))})$$ by currying and preservation of ends: $$\int_{A,B} \hom(F(A)\times\int^C \hom(J(A),C) \times \hom(C, J(B)) , F(B))$$ by Yoneda: $$\int_{A,B} \hom(F(A)\times \hom(J(A), J(B)), F(B))$$ by preservation of ends and fully-faithfulness of $J$: $$\int_{B} \hom(F(B), F(B))$$ $\endgroup$
    – Michal R. Przybylek
    Commented Apr 22, 2013 at 11:08
  • $\begingroup$ In the first sentence under the heading "The Object", you refer to $F(b)$. What does that mean, considering that $b$ is an object of $B$ while the domain of $F$ is only $A$? $\endgroup$
    – Andreas Blass
    Commented Apr 22, 2013 at 11:36
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    $\begingroup$ @Vidit and @Michal: There is an equivalent but less notation-heavy way to construct the morphism in question. If $f_!$ denotes left Kan extension along $f$, and $f_\ast$ denotes right Kan extension, then we have adjunctions $f_! \dashv f^\ast \dashv f_\ast$. Here, $L \dashv R$ means $L$ is left adjoint to $R$. From those adjunctions, we conclude that a morphism $f_! X \to f_\ast X$ as in the question is the same as (1) a morphism $X\to f^\ast f_\ast X$, and (2) a morphism $f^\ast f_! X \to X$. (to be continued) $\endgroup$
    – Ricardo Andrade
    Commented Apr 22, 2013 at 21:02
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    $\begingroup$ (continued) But these latter morphisms can be chosen as (1) the inverse of the counit of the adjunction $f^\ast \dashv f_\ast$, and (2) the inverse of the unit of the adjunction $f_! \dashv f^\ast$. Those morphisms are invertible as long as you assume that $f$ is full and faithful. I have not been able to prove abstractly that choices (1) and (2) lead to the same morphism $f_! X \to f_\ast X$, although one can check it explicitly, as Michal does, by using the description of Kan extensions as ends/coends. $\endgroup$
    – Ricardo Andrade
    Commented Apr 22, 2013 at 21:26

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