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 $[f,\mathcal{C},\mathcal{D}]$?

  2. Where can I get more information?

  • 1
    $\begingroup$ 2. What idea? You just write down a definition. There is no idea at all. Why do you want to consider this category of functors? Do you have an explicit example or application in mind? This should also answer 1. (there is no common notation because probably nobody cares). $\endgroup$ Commented Dec 14, 2013 at 16:32
  • $\begingroup$ I am not aware any of them, your notation looks fine. Anyway, for what would it be useful for you? $\endgroup$
    – Berci
    Commented Dec 14, 2013 at 16:33
  • $\begingroup$ @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 $X$... $\endgroup$ Commented Dec 14, 2013 at 16:38
  • $\begingroup$ ... 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. $\endgroup$ Commented Dec 14, 2013 at 16:43
  • $\begingroup$ @Berci, for motivation see the above two comments. $\endgroup$ Commented Dec 14, 2013 at 16:43

1 Answer 1


Trivial answer: It is the pullback $[C,D] \times_{[C',D]} 1$ where $f : 1 \to [C',D]$.


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