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Between two categories $\mathcal{C}$ and $\mathcal{D}$ there is a functor category $Fun(\mathcal{C},\mathcal{D})$ with functors as objects and natural transformations as morphisms.

Is there a universal property that can be used to define a functor category (up to equivalence)?

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What do you mean? Inside which category would you consider this universal property? –  Berci Jan 18 '13 at 13:00

3 Answers 3

up vote 2 down vote accepted

Here is a functor category that has a 2-universal property:

Theorem. Let $\mathbb{C}$ be a small category and let $h_\bullet : \mathbb{C} \to [\mathbb{C}^\textrm{op}, \textbf{Set}]$ be the Yoneda embedding. Then, for all locally small and cocomplete categories $\mathcal{E}$, the functor $F \mapsto F h_\bullet$ from the category of cocontinuous functors $[\mathbb{C}^\textrm{op}, \textbf{Set}] \to \mathcal{E}$ to the category of all functors $\mathbb{C} \to \mathcal{E}$ is fully faithful and essentially surjective on objects, and this functor is pseudonatural in $\mathcal{E}$. In other words, $[\mathbb{C}^\textrm{op}, \textbf{Set}]$ is the free cocompletion of $\mathbb{C}$.

But more generally, if $\mathbb{C}$ and $\mathbb{D}$ are both small categories, then $[\mathbb{C}, \mathbb{D}]$ has a 1-universal property:

Theorem. There is a bijection between functors $\mathbb{E} \times \mathbb{C} \to \mathbb{D}$ and functors $\mathbb{E} \to [\mathbb{C}, \mathbb{D}]$ and this bijection is natural in $\mathbb{C}, \mathbb{D}, \mathbb{E}$. In other words, the category of small categories is cartesian closed and the functor category is an exponential object.

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Thanks, do you have any reference on the second theorem? –  Michael Scarn Jan 18 '13 at 13:10
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No, but it's a straightforward calculation and analogous to the one for $\textbf{Set}$. –  Zhen Lin Jan 18 '13 at 14:16
    
IIRC this shows an argument for the second point (which really helped me grok why are natural transformations defined that way). Server is not responding at the moment, so I can't check :/ –  askyle May 29 at 15:24
    
OK, the link works now. Check section 7.6 (page 16). –  askyle May 29 at 15:49

Any category can be a functor category, because let $\mathcal C:=\{1\}$ the trivial category with one object, then $Fun(\mathcal C,\mathcal D)\simeq \mathcal D$.

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Sorry, I don't get your point. –  Michael Scarn Jan 18 '13 at 13:10
    
I didn't get yours either :) –  Berci Jan 18 '13 at 20:51

Amplifying Zhen's answer. One may show, that $[\mathbb{C}, \mathbb{D}]$ satisfies the 2-universal property, i.e. there exists a 2-natural isomorphism of categories: $$\hom(\mathbb{E} \times \mathbb{C}, \mathbb{D}) \approx \hom(\mathbb{E}, [\mathbb{C}, \mathbb{D}])$$ making $\mathbf{Cat}$ a 2-cartesian closed category. Moreover this structure is the unique 2-closed symmetric monoidal structure on $\mathbf{Cat}$.

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The amusing thing is that this can be proved using the 1-Yoneda lemma knowing only the 1-universal property. –  Zhen Lin Jan 18 '13 at 15:24
    
Can you elaborate on that a little? –  Berci Jan 18 '13 at 20:53
    
@Berci, every true statement can be proved using any assumptions (responding in line with your answer to the above question). However, I guess there is something more in Zhen's comment. Every cartesian closed category may be thought as enriched over itself. By definition, a 2-category is a category enriched over cartesian structure in $\mathbf{Cat}$. So, particularly, the 2-dimiensional structure on $\mathbf{Cat}$ is induced by its enrichment over itself. (cont) –  Michal R. Przybylek Jan 19 '13 at 12:37
    
This means that $\hom(\mathbb{A}, \mathbb{B})$ is really $[\mathbb{A}, \mathbb{B}]$ and $\hom(\mathbb{E} \times \mathbb{C}, \mathbb{D}) \approx \hom(\mathbb{E}, [\mathbb{C}, \mathbb{D}])$ turns into $[\mathbb{E} \times \mathbb{C}, \mathbb{D}] \approx [\mathbb{E}, [\mathbb{C}, \mathbb{D}]]$. The last condition is trivially true in any cartesian closed category by ordinary (1-dimensional) Yoneda lemma. I guess, by a similar argument, one may also show that in any cartesian closed category these isomorphisms are enriched natural, however, I have not checked all relevant detail. –  Michal R. Przybylek Jan 19 '13 at 12:38

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