Leinster question on isomorphic functor categories Hi so the question is basically to prove that for categories A and B
$$
[A,B]^{op} \simeq [A^{op},B^{op}]
$$
where these are functor categories.
So my first port of call was to write out what the structure of these beasts are and then to try and conjure up the relevant isomorphisms. Alas here I am.
First up I noted that 
$[A,B]^{op}$ has the same objects as $[A,B]$ , namely functors $F,G:A \rightarrow B$, but here the morphisms would be reversed i.e the natural transformations $\alpha:F\rightarrow G$ in $[A,B]$ become $\tilde{\alpha}:G\rightarrow F$ in $[A,B]^{op}$
Secondly for $[A^{op},B^{op}]$ I thought that for this category objects are functors $H,I:A^{op}\rightarrow B^{op}$ and morphisms are some natural transformations between them $\alpha':A^{op} \rightarrow B^{op}$ merrily mapping the inverse morphisms of $A$ and $B$ to each other.
So I feel like there is some deeper work to be done on this second point on relating it to $[A,B]$ in a concrete and explicit manner so an isomorphism could be explicitly constructed. This is the end of the road as far as my thought path goes however. I'm aware that these problems usually solve themselves on having a clear understanding of exactly what is entailed in the question; any help to achieving this would be greatly appreciated. Of course for bonus points the rigorous proof is the dream response.
 A: It may be easier to think that for arbitrary category $\mathcal{C}$ objects and morphisms of $\mathcal{C}^{op}$ are the same as objects and morphisms of $\mathcal{C}$, i.e. $Obj(\mathcal{C}^{op})=Obj(\mathcal{C})$ and $Mor(\mathcal{C}^{op})=Mor(\mathcal{C})$. The only difference between $\mathcal{C}$ and $\mathcal{C}^{op}$ is that all directions of morphisms of $\mathcal{C}^{op}$ are reversed. Note, that $(\mathcal{C}^{op})^{op}=\mathcal{C}$. With such approach, we get that objects and morphisms of $[A,B]^{op}$ are simply objects and morphisms of $[A,B]$. 
The next step is to point out that if $T\colon A\to B$ is a functor, then we can construct a functor $T^{op}\colon A^{op}\to B^{op}$, which is equal to $T$ on objects and morphisms. Note, that $(T^{op})^{op}=T$.
Next, for every natural transformation $\alpha\colon T\to S$ we can construct a natural transformation $\alpha^{op}\colon S^{op}\to T^{op}$, which is equal to $\alpha$ on objects of $A$. Note, that $(\alpha^{op})^{op}=\alpha$.
After that, the check that the assigning $T$ to $T^{op}$ and $\alpha$ to $\alpha^{op}$ is an isomorphism between $[A,B]^{op}$ and $[A^{op},B^{op}]$ becomes an easy exercise.
Hint. Let's denote the obtained functor by $\text{op}_{A,B}\colon [A,B]^{op}\to [A^{op},B^{op}]$. Try to prove that $$\text{op}_{A^{op},B^{op}}^{op}=\text{op}_{A,B}^{-1}.$$
Edit (rigorous proof). Let $A$ and $B$ be categories. Then define the functor $\text{op}_{A,B}\colon[A,B]^{op}\to[A^{op},B^{op}]$ in the following way: $\text{op}_{A,B}(T)=T^{op}$ and $\text{op}_{A,B}(\alpha)=\alpha^{op}$ for every $T\colon A\to B$ and every $\alpha\colon T\to S$, where $T,S\colon A\to B$. Note, that $\text{op}_{A,B}$ is indeed a functor, because $(\beta\circ\alpha)^{op}=\alpha^{op}\circ\beta^{op}$ and $1_{T}^{op}=1_{T^{op}}$ (see (*) for the proof). Note also that $\text{op}_{A,B}$ is an isomorphism of categories, because it has an inverse $\text{op}_{A^{op},B^{op}}^{op}$ (because for every category $A$, functor $T$ and natural transformation $\alpha$ we have $(A^{op})^{op}=A$, $(T^{op})^{op}=T$ and $(\alpha^{op})^{op}=\alpha$). 
(*) Let's prove that $(\beta\circ\alpha)^{op}=\alpha^{op}\circ\beta^{op}$. Indeed, for every object $a\in Obj(A)$ we have 
$$
(\beta\circ\alpha)^{op}(a)=(\beta\circ\alpha)(a)=\beta(a)\circ_B\alpha(a)=\alpha^{op}(a)\circ_{B^{op}}\beta^{op}(a)=(\alpha^{op}\circ\beta^{op})(a).
$$
Let's prove that $1_T^{op}=1_{T^{op}}$. Indeed,
$$
1_T^{op}(a)=1_T(a)=id_B(T(a))=id_{B^{op}}(T^{op}(a))=1_{T^{op}}(a).
$$
A: Look at what a morphism $\alpha:F\to G$ looks like in $[A^o,B^o]$. It is a collection $\{\alpha_X:FX\to GX \}$ of morphisms in $B^o$, i.e., a collection $\{\bar{\alpha}_X:GX \to FX\}$ of morphisms in $B$.
Therefore, if you want to identify $[A^o,B^o]$ and $[A,B]^o$, a good idea would be to send $\alpha$ to $\bar{\alpha}$. 
Now, you should verify that $\alpha$ is a natural transformation if and only if $\bar{\alpha}$ is.
