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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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    @QiaochuYuan: You're right, "always" should not be there. Thanks for noticing it.2012-11-10

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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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    Dear @Andŕe, you're welcome, I hoped as much!2012-11-10
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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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    This functor is *not* an isomorphism of categories! An isomorphism of categories is by definition given by a covariant functor. This functor is contravariant and therefore not considered an isomorphism of categories.2018-04-29
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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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    @Andre: Nope, because (in the formulation I'm referring to) by definition, $A \cdot_{\text{op}} B = B \cdot A$ (multiplication on the right is in $Mat$ and on the left is in $Mat^\circ$) and there exist $A$ and $B$ such that $B \cdot A \neq A \cdot B$. If $Mat^\circ = Mat$, then $\cdot$ and $\cdot_{\text{op}}$ would be the same operation.2012-11-11
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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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    @DávidTóth That's not an isomorphism. Because it is not a covariant functor.2018-04-29