Note: I've substantially edited the definition; $f$ is now allowed to be a functor.

Given categories $\mathcal{C}$ and $\mathcal{D}$, we can form the functor category $[\mathcal{C},\mathcal{D}]$. Now suppose we also have a functor $f : \mathcal{C}' \rightarrow \mathcal{D}$ where $\mathcal{C}'$ is a subcategory of $\mathcal{C}$. Then there is a subcategory of $[\mathcal{C},\mathcal{D}],$ which we can denote $[f,\mathcal{C},\mathcal{D}],$ defined as follows.

• Objects are precisely the functors $F : \mathcal{C} \rightarrow \mathcal{D}$ that agree with $f$, and
• Arrows are precisely the natural transformations $\nu : F \rightarrow G : \mathcal{C} \rightarrow \mathcal{D}$ such that $\nu(X) = \mathrm{id}_{f(X)}$ for all $X \in \mathrm{Ob}\,\mathcal{C}'$.

A few questions:

1. What is the proper notation for $[f,\mathcal{C},\mathcal{D}]$?

• @MartinBrandenburg, ah sorry I figured it was a common idea in common use. The particular motivation I have in mind is as follows. We can view a vector space as a model in $\mathrm{Set}$ of a two-sorted signature, one for scalars and one for vectors. Call the scalar sort $A$ and the vector sort $X$... – goblin Dec 14 '13 at 16:38
• ... Then natural transformations between models are homomorphisms. However, we're allowing non-identity arrows between the fields of scalars, which is non-standard. So to fix this, we can consider different subcategories of the category of models. A preliminary idea is to pick a "field" $F$ in $\mathrm{Set}$ and to let $f$ denote the function whose domain is $\{A\}$ with $f(A)=F$. Of course, this doesn't work because there are no fields in $\mathrm{Set}$! But I think that with a bit of cleverness, we should be able to choose an $f$ that gives the desired result. – goblin Dec 14 '13 at 16:43
Trivial answer: It is the pullback $[C,D] \times_{[C',D]} 1$ where $f : 1 \to [C',D]$.