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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?

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I don't understand what you mean by "always" here. "Opposite category" is a single functor $\text{Cat} \to \text{Cat}$ from the category of (say, small) categories to itself. –  Qiaochu Yuan Nov 10 '12 at 21:58
    
@QiaochuYuan: You're right, "always" should not be there. Thanks for noticing it. –  Andŕe Nov 10 '12 at 22:16
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4 Answers

up vote 4 down vote accepted

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.

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Not so silly :) –  Manos Nov 10 '12 at 21:57
    
@Andrew: Thanks, for giving an example helped me "see it". –  Andŕe Nov 10 '12 at 22:19
    
Dear @Andŕe, you're welcome, I hoped as much! –  Andrew Nov 10 '12 at 22:22
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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.

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Why contravariant? –  Berci Nov 10 '12 at 21:46
    
If $F$ is the functor $C \rightarrow C^{op}$, and $f:A \rightarrow B$ is a morphism in $C$, then $F(f)$ will be a morphism from $B=F(B)$ to $A=F(A)$. By definition of the opposite category, arrows from $A$ to $B$ in $C$ are viewed as arrows from $B$ to $A$ in $C^{op}$. –  Manos Nov 10 '12 at 21:52
    
Ahh... I thought, it is about the $()^{op}$ functor... –  Berci Nov 10 '12 at 21:59
    
Depending on foundations, $C^\circ$ also has the same morphisms as $C$ -- the difference is in the composition, source, and target operations. For the category corresponding to an abelian group, we would even have $C^\circ = C$ as a true equality. –  Hurkyl Nov 10 '12 at 22:18
    
@Hurkyl: and before posting this question, I was looking at $Mat$, the category of matrices. Just to be sure, in this case it is true that $Mat^{\text{op}} = Mat$, right? –  Andŕe Nov 10 '12 at 22:22
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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.

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  1. 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 $C^{op}\to D^{op}$.
  2. 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.
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