Opposite category

Given a category $\mathcal{C}$ , we may form another category $\mathcal{C}^{op}$ , called the opposite category or the dual category by 'reversing all the morphisms of $\mathcal{C}$ .

Formally we define $\mathcal{C}^{op}$ by:

$\text{Ob}(\mathcal{C}^{op}) = \text{Ob}(\mathcal{C})$
Every morphism $f:A\to B$ in $\mathcal{C}$ is associated to a unique morphism $f^{op}:B\to A$ in $\mathcal{C}^{op}$ (and every morphism $g$ in $\mathcal{C}^{op}$ is equal to $f^{op}$ for some $f$ in $\mathcal{C}$ ).
For any $f^{op}:B\to A$ and $g^{op}:C\to B$ the composition of $g^{op}$ and $f^{op}$ is definied by: $f^{op}\circ g^{op} = (g\circ f)^{op}.$

It is know easy to verify that $\mathcal{C}^{op}$ is a category:

Given $f^{op}:B\to A$ , $g^{op}:C\to B$ and $h^{op}:D\to C$ we have: $f^{op}\circ(g^{op}\circ h^{op}) = f^{op}\circ(h\circ g)^{op} = [(h\circ g)\circ f]^{op} = [h\circ (g\circ f)]^{op} = (g\circ f)^{op}\circ h^{op} = (f^{op}\circ g^{op})\circ h^{op}.$
For any $A\in \text{Ob}(\mathcal{C}^{op}) = \text{Ob}(\mathcal{C})$ we claim that the identity morphism on $A$ is just $1_A^{op}$ . Indeed, for any $f^{op}:B\to A$ , we have: $1_A^{op}\circ f^{op} = (f\circ 1_A)^{op} = f^{op} = (1_B\circ f)^{op} = f^{op}\circ 1_B^{op}.$

Duality Principle

The existence of the opposite category allows us to prove the duality principle which roughly states that: for any 'theorem' provable from the axioms of category theory, it's dual statement (i.e. the statement formed by 'replacing every morphism with a morphism pointing in the other direction') is also true (and can be proved by going through the exact same proof in the opposite category).

Similarly, given any construction or definition in category theory, we can from a dual concept in the same way (by defining that concept in the opposite category). The dual of a concept sometimes named by attaching the prefix co- to the name of the original concept (for instance, the dual of a limit is called a colimit).

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Opposite category

Duality Principle