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Given a functor $F : A \rightarrow B$, I managed to define the functor $F^{op} : A^{op} \rightarrow B^{op}$.

Now, I cannot see how to use it to define a functor from and to $\text{Cat}$ such that $C \longmapsto C^{op}$ and $F \longmapsto F^{op}$ for $C$ category and $F$ functor.

For instance, suppose that $\text{id}$ is the identity functor. Now we have to show: $\text{id}_c^{op} = \text{id}_{c^{op}}$. Do I proceed by case analysis?

I also have trouble when proving that $(F \circ G)^{op} = F^{op} \circ G^{op}$ because I arrive at something resembling a contra-variant functor.

This exercise is taken from Lecture Notes: Introduction to Categorical Logic

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1 Answer 1

The arrows of $A^{op}$ are the same but the source and target and composition are reversed. Thus, $F^{op}$ as a mapping is the same as $F$ (that is, $F^{op}(\alpha)=F(\alpha)$). In particular, $id^{op}$ is still the identity map, and the composition of functors is also clearly preserved.

$F^{op}$ is, how to say, contra-contravariant, is of the form $A^{op}\to B^{op}$, though a contravariant functor would be a (covariant) functor of the form $A^{op}\to B$ or, equivalently(!) $A\to B^{op}$.

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+1 for contra-contravariant :-) –  Julian Kuelshammer Nov 11 '12 at 23:49
    
Intuitively, the "id-condition" of the above functor is easy to see and, in the hunt for a formal justification, I already have a by case analysis (so ugly!) one. The "contra-contravariant" is the one can't see, even informally. Could you spare another hint on this? Thanks anyway. –  Andŕe Nov 12 '12 at 0:20

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