Every monadic functor is conservative

Let $f:(A,a)\to (B,b)$ be an algebra homomorphism whose underlying morphism has inverse $g$. From $f\circ a=b\circ T(f)$, the inverse equations, and functoriality, one obtains $g\circ b=a\circ T(g)$ by composing with $f$ and cancelling it; hence $g$ is an algebra homomorphism.

Therefore the Eilenberg–Moore forgetful functor reflects isomorphisms: an underlying inverse is automatically an inverse inside the algebra category by step 1.1.

For a monadic $U$, its comparison $K$ is an equivalence and hence fully faithful. If $U(f)=U^{T}K(f)$ is an isomorphism, step 2.1 makes $K(f)$ an isomorphism, and full faithfulness reflects that isomorphism back to $f$; thus $U$ is conservative.