Good day everyone:

I have been reading the book Categories for the Working Mathematicians and it is written that the functor category $B^{C}$ is itself a functor of the categories $B$ and $C$, covariant in $B$ and contravariant in $C$.

Specifically it is a functor $\mathsf{Cat}^{\mathrm{op}}\times \mathsf{Cat}\rightarrow \mathsf{Cat}$ which for objects

$$(C, B)\mapsto B^{C}$$

and for a pair of functors $F:B\rightarrow B'$ and $G:C'\rightarrow C$

$$(G,F)\mapsto F^{G}:B^{C}\rightarrow B'^{C'}$$

which is defined on objects $S\in B^{C}$ as $$F^{G}S=F\circ S\circ G$$ and for a natural transformation $\tau:S\rightarrow T$ in $B^{C}$, $$F^{G}\tau=F\circ\tau\circ G.$$

My question is why is necessary $\mathsf{Cat}^{\mathrm{op}}$ ? or why does $C$ have to be contravariant the functor?

Thank you for your time.

  • $\begingroup$ I've posted a response to your (now deleted) comment. $\endgroup$ Jul 7, 2015 at 20:59

1 Answer 1


Suppose you have a functor $T:C_1\to C_2$. The only "obvious" map of functor categories available, namely composing with $T$, is contravariant, i.e. it is a map $B^{C_2}\to B^{C_1}$: $$(C_2\xrightarrow{\;F\;} B)\in B^{C_2}\qquad \leadsto \qquad (C_1\xrightarrow{\;T\;}C_2\xrightarrow{\;F\;}B)\in B^{C_1}$$ How do you propose defining a map $B^{C_1}\to B^{C_2}$?

  • $\begingroup$ The "functor category" functor (let's call it $\mathsf{Func}$) is defined as taking a pair $(C,B)$ to the category $B^C$. So what $\mathsf{Func}$ does to the objects $(C,B)$ is fixed. If you want $\mathsf{Func}$ to be covariant in $C$, then your goal is to define the functor $$\mathsf{Func}:\mathsf{Cat}\times\mathsf{Cat}\to\mathsf{Cat}$$ $$\mathsf{Func}(C,B)=B^C$$ such that $\mathsf{Func}$ takes a map $$(C_1,B)\xrightarrow{\;\;(T,\mathrm{id}_B)\;\;}(C_2,B)$$ to a map $$B^{C_1}\xrightarrow{\;\;\mathsf{Func}(T,\mathrm{id}_B)\;\;} B^{C_2}$$ $\endgroup$ Jul 7, 2015 at 20:57
  • $\begingroup$ What is your proposal for defining the rule $\mathsf{Func}(T,\mathrm{id}_B)$ for taking an element $F\in B^{C_1}$ and sending it to an element of $B^{C_2}$? $\endgroup$ Jul 7, 2015 at 20:59
  • $\begingroup$ i was thinking about taking $Dual (T):C_{2}\rightarrow C_{1}$ however i don't think if i can do it because i don't know if Dual(T) exists. $\endgroup$
    – Liddo
    Jul 7, 2015 at 21:06
  • $\begingroup$ i would use $T^{op}$ and with that $F\circ T^{op}$ does the job $\endgroup$
    – Liddo
    Jul 7, 2015 at 22:28
  • $\begingroup$ Except what does "$T^{\mathrm{op}}$" mean? While it's true you can consider $T$ as an element $$\large T\in\mathrm{Hom}_{\mathsf{Cat}^{\mathrm{op}}}(C_2,C_1)$$ that doesn't make it a functor from $C_2$ to $C_1$, and even if it did do that, you can't make the composition $F\circ T^{\mathrm{op}}$ because $F$ is a functor $$F:C_2\longrightarrow B$$ $\endgroup$ Jul 7, 2015 at 22:32

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