Assume the equivalent conditions and that $T$ preserves finite coproducts. The free functor already reflects isomorphisms. By the dual <Crude monadicity theorem>, it remains to preserve the relevant coreflexive equalizers.

A coreflexive equalizer diagram
$$
A\longrightarrow B\rightrightarrows C
$$
with $A\neq\varnothing$ can be equipped with the extra sections making it a split equalizer: choose a point of $A$ and use the common retraction to define the missing splitting maps on the complementary fibres. Every functor preserves split equalizers. If $A=\varnothing$, preservation follows from preservation of the initial object, which follows from preservation of finite coproducts. Consequently $F$ preserves all required coreflexive equalizers, and the adjunction is <comonadic adjunction>[comonadic].

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 119 / 4 / c / Solution

{wiki=Beck's_monadicity_theorem}

An adjunction is comonadic when the comparison from its left-hand category to coalgebras for the induced comonad is an equivalence. The dual Beck theorem tests this using reflected isomorphisms and suitable equalizers.


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 Incoming links: Comonadic adjunction