Units, Counits & Triangle Identities

Feed the identity arrows into the adjunction bijection and two natural transformations fall out; the triangle identities say that a generator, read as a one-letter word and multiplied out, is the generator again.

The idea

The unit and the counit repackage an adjunction as two natural transformations, which together with the triangle identities carry exactly the same information as the bijection that defined it.

An adjunction $F \dashv G$ is a bijection $\mathcal{D}(F(A), B) \cong \mathcal{C}(A, G(B))$, natural in both variables — one bijection for every pair of objects, which is awkward to compute with. The smaller package comes from transposing the identity arrows. Write $\overline{g}$ for the arrow matched with $g$, in either direction.

Definition (Unit and counit).

Let $F \dashv G$. Since $1_{F(A)}$ lies in $\mathcal{D}(F(A), F(A))$ and $1_{G(B)}$ lies in $\mathcal{C}(G(B), G(B))$, transposing them gives arrows $\eta_{A} = \overline{1_{F(A)}} \colon A \to GF(A), \qquad \varepsilon_{B} = \overline{1_{G(B)}} \colon FG(B) \to B,$ called the unit and the counit of the adjunction.

Theorem (Triangle identities).

The unit and counit are natural transformations $\eta \colon 1_{\mathcal{C}} \Rightarrow GF$ and $\varepsilon \colon FG \Rightarrow 1_{\mathcal{D}}$, and for every object $A$ of $\mathcal{C}$ and $B$ of $\mathcal{D}$ they satisfy $\varepsilon_{F(A)} \circ F(\eta_{A}) = 1_{F(A)}, \qquad G(\varepsilon_{B}) \circ \eta_{G(B)} = 1_{G(B)}.$

In the free-against-forgetful adjunction between sets and groups, the unit sends a generator to the one-letter word and the counit multiplies a formal word out inside the group. The first identity then reads: regard a generator as a word of length one, multiply it out, and the generator comes back. The pattern is general — the unit inserts the generators, and the counit evaluates.

Ways to work on it

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