FALSE: ordinary Beck creation characterizes strict monadicity

Let $\mathcal{D}$ have objects tagged sets $(X,i)$ with $i\in \{0,1\}$ and let every morphism $(X,i)\to (Y,j)$ be a function $X\to Y$. The forgetful functor $U:\mathcal{D}\to \mathbf{S}\mathbf{e}\mathbf{t}$ has a left adjoint $F(X)=(X,0)$.

The unit and counit of this adjunction act as identity functions, so the induced monad on $\mathbf{S}\mathbf{e}\mathbf{t}$ is the identity monad.

Its object map is not injective because $(X,0)$ and $(X,1)$ are distinct objects with the same image. Hence the comparison is not an isomorphism by [L3] and $U$ is not strictly monadic.

The comparison is $U$ itself. It is fully faithful because every function $X\to Y$ is a morphism $(X,i)\to (Y,j)$ and no two morphisms have the same underlying function, and the assignment $X\mapsto (X,0)$ splits it on objects, since $U(X,0)=X$ on the nose. By [L4] it is therefore an equivalence, so $U$ is monadic in the sense of [L1].

By [L2], ordinary Beck creation holds for this monadic right adjoint, while step 2.2 shows strict monadicity fails. Thus the implication from ordinary creation to strict monadicity is false.

The other implication is true: strict monadicity implies monadicity by [L1], and monadicity implies ordinary creation by [L2]. Hence step 4.1 refutes exactly the reverse implication in the displayed biconditional; strict creation is the additional condition characterized by strict Beck.