# Transferring properties from a category $\mathcal{D}$ to the functor category $[\mathcal{C}, \mathcal{D}]$

Often in category theory, a property of a category $\mathcal{D}$ will also hold in the functor category $[\mathcal{C}, \mathcal{D}]$, perhaps with some smallness conditions on $\mathcal{C}$. The proofs all follow the same pattern: take a natural transformation $\alpha : F \to G$, consider each component $\alpha_C : FC \to GC$ separately — so that we're now working in $\mathcal{D}$ — use the property of $\mathcal{D}$ and then patch the results together at the end.

For instance:

• If $\mathcal{D}$ has (co)products then so does $[\mathcal{C}, \mathcal{D}]$.
• If $\mathcal{D}$ is abelian (or additive, or preadditive, or...) then so is $[\mathcal{C}, \mathcal{D}]$, provided $\mathcal{C}$ is locally small.

Question: Are there any general results, such as conditions on a given property $\phi$, that ensure that if $\phi$ holds in $\mathcal{D}$ then $\phi$ holds in $[\mathcal{C},\mathcal{D}]$.

My thoughts: The answer is inevitably going to have something to do with the 'patching together' part of the proof.

For example, if $\mathcal{D}$ has products and $F, G : \mathcal{C} \rightrightarrows \mathcal{D}$ then we can define the product $F \times G : \mathcal{C} \to \mathcal{D}$ by $(F \times G)C = FC \times GC$, and then the projection arrows are defined componentwise by $$(\pi_F)_C = \pi_{FC} : FC \times GC \to FC$$ $$(\pi_G)_C = \pi_{GC} : FC \times GC \to GC$$ The fact that $(F \times G, \pi_F, \pi_G)$ defines a product in $[\mathcal{C}, \mathcal{D}]$ hangs on the fact that the componentwise definition really does yield a natural transformation, rather than just an $\operatorname{ob} \mathcal{C}$-indexed family of arrows.

• Being preadditive is a structure, not a property. (Somewhat surprisingly, though, being additive is a property.) – Qiaochu Yuan Jan 5 '13 at 11:56
• @QiaochuYuan: I'm just as interested in structure as I am in properties. But the title's already quite long! – Clive Newstead Jan 5 '13 at 12:00
• @QiaochuYuan, where can I find definitions of structure and property? And are they universally agreed upon? – alancalvitti Jan 6 '13 at 4:12

Cilve, your first instance shows that we have to be careful when moving from $\mathcal{D}$ to $[\mathcal{C}, \mathcal{D}]$ --- you should easily find categories $\mathcal{C}, \mathcal{D}$ such that $\mathcal{D}$ is cartesian closed, but the functor category $[\mathcal{C}, \mathcal{D}]$ is not.

The other examples follow more-or-less from a suitable version of the Yoneda lemma. I will show you how to apply Yoneda lemma to get limits/colimits in a functor category. Let $\mathcal{X}, \mathcal{D}$ be small categories. There exists their cotensor (exponent) $\mathcal{D}^\mathcal{X}$ together with the diagonal functor: $$\mathcal{D} \overset{\Delta}{\rightarrow} \mathcal{D}^\mathcal{X}$$ given as the transposition of the cartesian projection $\pi \colon \mathcal{D} \times \mathcal{X} \rightarrow \mathcal{D}$. The limit functor $\mathit{lim}$ is defined as the right adjoint to the diagonal, and the colimit $\mathit{colim}$ is defined as the left adjoint to the diagonal.

The whole idea is to apply the 2-Yoneda functor to the above diagram. We have $$\hom(-, \mathcal{D}^\mathcal{X}) \approx \hom(-, \mathcal{D})^{\mathcal{X}}$$ by the definition of the cotensor. Therefore the above diagram is mapped to the diagram: $$\hom(-, \mathcal{D}) \rightarrow \hom(-, \mathcal{D})^\mathcal{X}$$ Since adjunctions are equationally defined, they are preserved by any 2-functor, and particularly by 2-Yoneda. This means, that the above transformation has right/left adjoint transformation provided $\Delta$ has. But a transformation that has left/right adjoint, has left/right adjoint on its each componenet $\mathcal{C}$. Thus: $$\hom(\mathcal{C}, \mathcal{D}) \rightarrow \hom(\mathcal{C}, \mathcal{D})^\mathcal{X}$$ has right/left adjoint if $\mathcal{D}$ has $\mathcal{X}$-indexed limits/colimits.

We may also see what would go wrong if one tried to apply the above strategy to show that cartesian closedness is inherited by functor categories.

Let us recall that a category $\mathcal{D}$ is cartesian closed if for every global element $x \colon 1 \rightarrow \mathcal{D}$ the canonical functor: $$\mathcal{D} \approx \mathcal{D} \times 1 \overset{\mathit{id}\times x}\rightarrow \mathcal{D} \times \mathcal{D} \overset{\times_\mathcal{D}}{\rightarrow} \mathcal{D}$$ has right adjoint, where $\mathcal{D} \times \mathcal{D} \overset{\times_\mathcal{D}}{\rightarrow} \mathcal{D}$ is the internal cartesian product functor in $\mathcal{D}$.

Just like before, we may apply to our diagram the 2-Yoneda functor obtaining: $$\hom(-, \mathcal{D}) \approx \hom(-, \mathcal{D}) \times 1 \overset{\mathit{id}\times \hom(-, x)}\rightarrow \hom(-, \mathcal{D}) \times \hom(-, \mathcal{D}) \overset{\times_{\hom(-, \mathcal{D})}}{\rightarrow} \mathcal{D}$$ and conclude that this transformation has right adjoint iff the former has. However, the terminal object $1$ is not a (2-)generator in $\mathbf{Cat}^{\mathbf{cat}^{op}}$, thus the adjunctions do not give a good characterisation of internally cartesian closed objects in that category. Particularly, $\hom(-, \mathcal{D}) \colon \mathbf{cat}^{op} \rightarrow \mathbf{Cat}$ may not be a cartesian closed (2-)fibration, and one may expect existence of exponents in a fibre $\hom(\mathcal{C}, \mathcal{D})$ only on "constant" objects induced by global sections $\hom(\mathcal{C}, x) \colon 1 \rightarrow \hom(\mathcal{C}, \mathcal{D})$.

• Yikes, I don't know why I thought that about Cartesian closed categories. I've edited my question, and I'll read the rest of your answer now! – Clive Newstead Jan 5 '13 at 12:38
• Finally got round to reading it properly. This is great, thanks a lot! – Clive Newstead Jan 7 '13 at 13:38