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I'm currently working through Tom Leinster's Basic Category Theory, and I have searched the internet fruitlessly for examples of functors between functor categories. I haven't yet come up with any intuition on how one might construct or characterize a functor $F : [\mathcal{A}, \mathcal{B}] \to [\mathcal{A}', \mathcal{B}']$, especially with the functor categories $[\mathcal{A}, \mathcal{B}]$ and $[\mathcal{A}', \mathcal{B}']$ as objects of CAT. I feel very bogged down in the details of how the functor $F$ sends functors to functors and natural transformations to natural transformations; I understand that depending on the scenario there might be a natural choice for $F$ mapping $f \in [\mathcal{A}, \mathcal{B}]$ to another functor, and similarly for the natural transformations, but that choice would have to be consistent with the functor and natural transformation axioms for each morphism and object in $\mathcal{A}$. I suppose my confusion is in the multiple layers of information that such a functor $F$ must encode, and how to keep this information consistent and organized when working with functors such as these.

For context, I'm trying to show that $[\mathcal{A^{\text{op}}}, \mathcal{B^{\text{op}}}] \cong [\mathcal{A}, \mathcal{B}]^{\text{op}}$ as objects of CAT, so I need to find two functors $F,G$ so that $FG = 1_{[\mathcal{A}, \mathcal{B}]^{\text{op}}}$ and $GF = 1_{[\mathcal{A^{\text{op}}}, \mathcal{B^{\text{op}}}] }$. I also have terrible intuition for opposite categories, and I cannot see a natural choice for $F$ or $G$. Every attempt I've made ends up with a construction of $F$ that is contravariant, which I don't think can give me an isomorphism.

Any intuition you can impart about my first question, and any hints as to the second would be appreciated.

Thanks.

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    $\begingroup$ A contravariant functor can be an isomorphism! $\endgroup$
    – Mathmo123
    Commented Dec 28, 2014 at 18:28
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    $\begingroup$ @Mathmo123 Wouldn't it be an anti-isomorphism? $\endgroup$ Commented Dec 28, 2014 at 20:14
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    $\begingroup$ For a trivial example consider the functor $ G-\mathbf {Set}\longrightarrow G'-\mathbf {Set} $ induced by a group homomorphism $ G'\longrightarrow G $. This can be generalized by replacing $ G, G',\mathbf {Set} $ by arbitrary categories. $\endgroup$
    – user158047
    Commented Dec 28, 2014 at 20:54
  • $\begingroup$ @JakobWerner That's very nice and simple. It doesn't appear to apply directly to the problem I'm considering, but that is a neat arrangement. Thanks. $\endgroup$ Commented Dec 28, 2014 at 21:48

1 Answer 1

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You want to define a contravariant functor $[\mathcal{A}^{op}, \mathcal{B}^{op}] \to [\mathcal{A},\mathcal{B}]$.

Given a functor $T : \mathcal{A}^{op} \to \mathcal{B}^{op}$, define $T_\ast : \mathcal{A} \to \mathcal{B}$ on objects by $T_\ast(X) = X$. Given a morphism $f:X\to Y$ of $\mathcal{A}$, view it as a morphism $f^{op} : Y \to X$ in $\mathcal{A}^{op}$; then $T(f^{op}) : T(Y) \to T(X)$ is a morphism in $\mathcal{B}^{op}$ which corresponds to some morphism $T(f^{op})^{op} : T(X) \to T(Y)$ in $\mathcal{B}$.

So we define $T_\ast$ on morphisms by setting $T_\ast(f) = T(f^{op})^{op}$.

You now need to check that $T_\ast$ is indeed a functor $\mathcal{A} \to \mathcal{B}$.

Assuming this is done, you have a rule that associates to each object $T \in [\mathcal{A}^{op}, \mathcal{B}^{op}]$ an object $T_\ast \in [\mathcal{A},\mathcal{B}]$.

Let's now define $()_\ast$ on morphisms in $[\mathcal{A}^{op},\mathcal{B}^{op}]$, i.e. natural transformations.

Given $\eta : T \to S$ a natural transformation between $T, S : \mathcal{A}^{op} \to \mathcal{B}^{op}$, we want to define a natural transformation $\eta_\ast : S_\ast \to T_\ast$ between $S_\ast, T_\ast : \mathcal{A} \to \mathcal{B}$.

Let $X$ be an object in $\mathcal{A}$. Then $\eta_X : T(X) \to S(X)$ is an arrow in $\mathcal{B}^{op}$, so $\eta_X^{op} : S(X) \to T(X)$ is an arrow in $\mathcal{B}$.

So we define $\eta_{\ast,X} : S_\ast(X) \to T_\ast(X)$ by the rule $\eta_{\ast,X} = \eta_X^{op}$.

You should now check that $\eta_\ast$ really is a natural transformation $S_\ast \to T_\ast$, and that $(\xi \circ \eta)_\ast = \eta_\ast \circ \xi_\ast$ in $[\mathcal{A},\mathcal{B}]$ whenever $\xi : S \to U$ is some other morphism in $[\mathcal{A}^{op}, \mathcal{B}^{op}]$.

Then you have a contravariant functor $()_{\ast} : [\mathcal{A}^{op}, \mathcal{B}^{op}] \to [\mathcal{A}, \mathcal{B}]$.

Applying this construction to the pair of categories $(\mathcal{A}^{op}, \mathcal{B}^{op})$ instead of $(\mathcal{A}, \mathcal{B})$ gives you a contravariant functor $()^{\ast} : [\mathcal{A}, \mathcal{B}] \to [\mathcal{A}^{op}, \mathcal{B}^{op}]$.

You should now check that $()_\ast$ and $()^{\ast}$ are mutually inverse equivalences of categories.

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  • $\begingroup$ Thanks. I'll work through the details. $\endgroup$ Commented Dec 28, 2014 at 23:02

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