 Solution

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 diagramwith $A\ne \varnothing$ 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.