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Composing with an adjunction that induces the identity monad on the nose does not change the induced monad

Statement

Let $F : C ⇄ D : G$ and $F^{'} : D ⇄ E : G^{'}$ be adjunctions. Suppose the monad induced by $F^{'} ⊣ G^{'}$ is the identity monad on $D$ on the nose: $G^{'} F^{'} = 1_{D}$ , $η^{'} = 1$ , and $G^{'} ε^{'} F^{'} = 1$ . Then the monad induced on $C$ by the composite adjunction $F^{'} F ⊣ G G^{'}$ is the monad induced by $F ⊣ G$ on the nose.

Facts & Assumptions

Given: The two adjunctions and the displayed strict identity-monad hypotheses.

[L1]

The composite adjunction has unit $\overset{ˉ}{η} = (G η^{'} F) \circ η$ and counit $\overset{ˉ}{ε} = ε^{'} \circ (F^{'} ε G^{'})$ (Adjunctions compose with the composite unit and counit formulas).

Proof

technique · direct

1.1

The composite induced endofunctor is $G G^{'} F^{'} F = G F$ , and [L1] with $η^{'} = 1$ gives $\overset{ˉ}{η} = η$ .

2.1

Its multiplication is $G G^{'} \overset{ˉ}{ε} F^{'} F$ . Expanding $\overset{ˉ}{ε}$ by [L1], then using $G^{'} F^{'} = 1_{D}$ and $G^{'} ε^{'} F^{'} = 1$ , reduces this transformation to $G ε F$ .

step 1.1L1

3.1

The endofunctor, unit, and multiplication obtained in steps 1.1 and 2.1 are exactly $G F$ , $η$ , and $G ε F$ , so the two induced monads agree on the nose.

step 1.1step 2.1∎

Depends on

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Sources

S. Mac Lane, Categories for the Working Mathematician, 2nd ed., Exercise VI.5.5 (standard reference, not scraped)