Duality & Opposite Categories

Reverse every arrow and every theorem comes with a second one, proved for free.

The idea

Definition (Opposite category).

The opposite $\mathcal{C}^{\mathrm{op}}$ of a category $\mathcal{C}$ has the same objects as $\mathcal{C}$; an arrow $A \to B$ of $\mathcal{C}^{\mathrm{op}}$ is an arrow $B \to A$ of $\mathcal{C}$, so $\mathcal{C}^{\mathrm{op}}(A,B) = \mathcal{C}(B,A)$; its identities are those of $\mathcal{C}$; and two arrows $f$ and $g$ whose composite in $\mathcal{C}$ is $g \circ f$ compose in $\mathcal{C}^{\mathrm{op}}$ to $f \circ g$.

So the opposite is obtained by turning every arrow around, and reversing twice puts everything back. The axioms survive the reversal — the identities were never touched, and associativity read backwards is still associativity — so $\mathcal{C}^{\mathrm{op}}$ is again a category. That observation is the duality principle. Any definition written in the language of categories can be read in $\mathcal{C}^{\mathrm{op}}$ instead of $\mathcal{C}$, and what comes out is a second, usually different, definition: its dual. Any theorem proved for all categories holds in particular for every opposite, so its dual is proved at the same moment, and dual notions come in pairs marked by a co- prefix: product and coproduct, limit and colimit.

The reversal produces genuinely new structure. A monoid is a category with one object, its elements being the arrows and its multiplication the composition. The opposite of the monoid of all functions $\{1,2,3\} \to \{1,2,3\}$ has the same elements but multiplies them in the other order — and composition of functions does not commute, so it is a different monoid on the same elements.

Ways to work on it

• Walkthrough. Building the opposite category, dualizing a statement, and why contravariant functors are ordinary functors.
• Practice. Produce the dual of a definition, count arrows in the opposite of an ordered set, and multiply in an opposite monoid.
• Hardest. Harvest a theorem for free by dualizing, then find a monoid that is not isomorphic to its own opposite.

Not sure where to start? Take the ten-question placement test.