1. ## Category Theory

What are $\displaystyle \mathcal{AB}^{op}$ and $\displaystyle \mathcal{GRP}^{op}$ naturally equivalent to?

Okay I'm stuck in this one, I know what a natural equivalence is, but my problem is I don't really understand what $\displaystyle (.)^{op}$ does, I know it reverses all arrows in the Category but I don't get what changes by doing that.

2. Originally Posted by Jose27
What are $\displaystyle \mathcal{AB}^{op}$ and $\displaystyle \mathcal{GRP}^{op}$ naturally equivalent to?

Okay I'm stuck in this one, I know what a natural equivalence is, but my problem is I don't really understand what $\displaystyle (.)^{op}$ does, I know it reverses all arrows in the Category but I don't get what changes by doing that.
In wiki (link), an opposite category or dual category of $\displaystyle C^{op}$ of a given category C is formed by reversing the morphisms. It means the objects of $\displaystyle C^{op}$ are just the objects of C and the arrows of $\displaystyle C^{op}$ are arrows $\displaystyle f^{op}$ in a one to one correspondence manner (if an arrow of C is $\displaystyle f:x \rightarrow y$, then $\displaystyle f^{op}$ in $\displaystyle C^{op}$ is $\displaystyle f^{op}:y \rightarrow x$. I assume you already know this one.

Now consider some examples. Let S be a functor from the category of $\displaystyle C^{op}$ to category of B such that $\displaystyle S:C^{op} \rightarrow B$. It assigns to each object $\displaystyle c \in C^{op}$ an object Sc of B and to each arrow $\displaystyle f^{op}:b \rightarrow a$ of $\displaystyle C^{op}$ an arrow $\displaystyle Sf^{op}:Sb \rightarrow Sa$ of B.
Here is another example. If $\displaystyle T:C \rightarrow B$ is a functor from the category of C to the category of B, then we can define a functor from $\displaystyle T^{op}:C^{op} \rightarrow B^{op}$. In this case, $\displaystyle c \mapsto Tc$ can be applied to both, where c is an object of both C and C^{op} (Remember that objects remain same in the dual category). However, $\displaystyle f \mapsto Tf$ should be changed to $\displaystyle f^{op} \mapsto (Tf)^{op}$ for the latter one.

3. Originally Posted by Jose27
What are $\displaystyle \mathcal{AB}^{op}$ and $\displaystyle \mathcal{GRP}^{op}$ naturally equivalent to?
this question is not easy to answer! right now i have no idea what category $\displaystyle \mathcal{GRP}^{op}$ is equivalent to but it is known that $\displaystyle \mathcal{AB}^{op}$ is equivalent to the category of compact abelian groups (continuous

homomorphisms are the morphisms of this category). the idea is to associate to any abelian group $\displaystyle G$ the abelian group of its characters, i.e. $\displaystyle \hat{G}=\text{Hom}_{\mathbb{Z}}(G, \mathbb{T}),$ where $\displaystyle \mathbb{T}=\mathbb{R}/\mathbb{Z}$ is the circle group.

we also associate to any homomorphism $\displaystyle f: G_1 \longrightarrow G_2$ the morphism $\displaystyle \hat{f}: \hat{G_2} \longrightarrow \hat{G_1}$ defined by $\displaystyle \hat{f}(g)=g \circ f.$ the topology on $\displaystyle \hat{G}$ is the pointwise convergence topology on $\displaystyle \mathbb{T}^G,$ where $\displaystyle \mathbb{T}^G$ is the set

of all functions $\displaystyle G \longrightarrow \mathbb{T}.$ there are so much details here to be proved and you probably can find find them in some textbooks.