Opposite category functor I need some confirmation: is the opposite category transformation always a functor?
Also, isn't it always the case that $C^{\text{op}} = C$, since the the way we label an arrow does not matter?
 A: The objects of $\mathcal C^{\text{op}}$ are exactly the same as the objects of $\mathcal C$.
The morphisms of $\mathcal C^{\text{op}}$ are backwards versions of the ones in $\mathcal C$.
If we have a morphism $A \color{green}{\longrightarrow} B$ in $C$, we have a morphism $B \color{blue}{\longrightarrow} A$ in $\mathcal C^{\text{op}}$.
These are different categories because there might be a morphism $A \color{green}{\longrightarrow} B$ in $C$ but no morphism $A \color{blue}{\longrightarrow} B$ in $C^{\text{op}}$.
There is a contravariant functor $$ \begin{array}{rrcl} &\text{Dual}&:& \mathcal C \longrightarrow \mathcal C^{\text{op}} \\ &\text{Dual}(A)&=& A \\ &\text{Dual}(A \overset{f}{\color{green}\longrightarrow} B)&=& B \overset{f}{\color{blue}{\longrightarrow}} A \end{array}$$ which is an isomorphism of categories though.
A: Yes, going from $C$ to $C^{op}$ is a contra-variant functor.
The equality $C^{op}=C$ is not true. Don't be misled by the fact that these two categories have exactly the same objects. The morphisms differ, and recall that a category is not simply the collection of objects, rather the objects together with the morphisms. 
A: *

*Yes, there is a functor $Cat\to Cat$ that maps $C\mapsto C^{op}$ and maps a functor $F:C\to D$ to the corresponding $C^{op}\to D^{op}$.

*No. Direction does matter. Well, how to say, it is easy to figure out $C^{op}$ once we know $C$, but these are not equivalent, neither identical.

A: Here is a silly example. Form a category with objects $a,b,c$ and morphisms $f:a\to b,g:a\to c$ (and identities). This category has an initial object, namely $a.$ On the other hand, its opposite category clearly does not have an initial object ($a$ becomes terminal). Thus the two categories must be distinct.
