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Let $\mathcal{C},\mathcal{D}$ be categories and $D\in\mathcal{D}$ a generator. (This means, that $\operatorname{Hom}_{\mathcal{D}}(C,\_)$ is faithful). The functor category $\operatorname{Fun}(\mathcal{C},\mathcal{D})$ posseses then a generator, too. How does it look like?

My idea was taking the constant functor $\equiv D\in\operatorname{Fun}(\mathcal{C},\mathcal{D})$. I am on the right way?

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1 Answer 1

The claim is false as stated. Consider $\mathcal{D} = \mathbf{Set}$: this is generated by $1$ in a very strong sense. However, if $\mathcal{C} = 2$ is the discrete category on two objects (say $a$ and $b$), then $[2, \mathbf{Set}]$ does not have a generator. Suppose for a contradiction that $G$ generates $[2, \mathbf{Set}]$. Clearly, $G (a)$ and $G (b)$ must be non-empty. However, if $H$ is an object in $[2, \mathbf{Set}]$ such that $H (a) = \emptyset$ and $H (b) = 1$, then there are no morphisms $G \to H$. Hence $G$ is not a generator.


Here is something that is true. If $\mathcal{C}$ is small and $\mathcal{D}$ is cocomplete and has a generator, then $[\mathcal{C}, \mathcal{D}]$ has a small generating family. The way to see this is to observe that the functor $F \mapsto F c : [\mathcal{C}, \mathcal{D}] \to \mathcal{D}$ has a left adjoint $c_! : \mathcal{D} \to [\mathcal{C}, \mathcal{D}]$, and using adjointness, if $G$ is a generator in $\mathcal{D}$, then the family $\{ c_! G : c \in \operatorname{ob} \mathcal{C} \}$ is a small generating family in $[\mathcal{C}, \mathcal{D}]$.

Moreover, if $\mathcal{D}$ is pointed, then $[\mathcal{C}, \mathcal{D}]$ has a generator: just take the coproduct $\coprod_{c \in \operatorname{ob} \mathcal{C}} c_! G$.

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Thank you. How can I fix this claim? Does it help, if $\mathcal{D}$ is abelian? –  Jack Jul 10 '13 at 10:28
    
No. You can replace $\mathbf{Set}$ with $\mathbf{Ab}$ in the above and run the same argument. –  Zhen Lin Jul 10 '13 at 10:35
    
Wikipedia says: "Given a small category C and a Grothendieck category A, the functor category Funct(C,A) is a Grothendieck category; if C is preadditive, then the functor category Add(C,A) of all additive functors from C to A is a Grothendieck category as well." So this is wrong? –  Jack Jul 10 '13 at 10:39
    
Hm, sorry – the argument doesn't work with $\mathbf{Ab}$. The first claim in your quote is correct – see my edit. –  Zhen Lin Jul 10 '13 at 11:03

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