Monadic functor

Remarks

A functor without a left adjoint is never monadic. Some authors call $U$ strictly monadic when $K$ is an isomorphism of categories. The comparison functor always satisfies $U^T \circ K = U$, where $U^T$ is the forgetful functor of $\mathcal{C}^T$.

Example: the forgetful functor from groups to sets is monadic. Its left adjoint is the free group functor. An algebra of the monad $T$ is a set $A$ with a map from the free group on $A$ to $A$ satisfying the algebra laws. Such a map comes from a unique group structure on $A$, and algebra morphisms are group homomorphisms, so $K$ is an isomorphism.

Non-example: the forgetful functor $U$ from topological spaces to sets has the discrete space functor as left adjoint. The monad $U F$ is the identity monad, whose algebras are just sets, so $\mathcal{C}^T$ is the category of sets. The comparison functor sends a space to its underlying set, and it is not full, since not every function between spaces is continuous. So $K$ is not an equivalence and $U$ is not monadic.