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Every order-preserving function from a complete lattice to itself has a least and a greatest fixed point. More strongly, its set of fixed points is itself a complete lattice.

Use $\mathbb{N}=\{0,1,\dots \}$, as required by the formula at zero, and composition of functions as multiplication in $M$. The displayed $T$ preserves identities, and for $n>0$,both sides send zero to zero. Thus $T$ is a functor on the one-object category.

For the shift monad on order-preserving maps of natural numbers, setThese are order-preserving functions. The equations $Tf\eta =\eta f$ and $\mu T^{2}f=Tf\mu$ hold pointwise, giving the required natural transformations. For the second equation, both sides are zero at $n=0,1$, and are $f(n-2)+1$ for $n\ge 2$. The two unit laws are $\mu \eta =\mu T\eta =1$. Associativity is checked byConsequently these maps define a monad. It is not an idempotent monad, since $\mu (0)=\mu (1)$, so $\mu$ cannot be invertible.

An algebra for a monad is a map $a\in M$ with $a\eta =1$ and $aTa=a\mu$. The first equation forces $a(n+1)=n$; monotonicity then forces $a(0)=0$. Thus $a=\mu$, which satisfies the second equation by the monad associativity law. The Eilenberg-Moore category therefore has exactly one object. Its endomorphisms $h$ satisfy $h\mu =\mu Th$, and this equation holds precisely when $h(0)=0$: evaluate at zero for necessity, and at $n>0$ both sides equal $h(n-1)$.

The Kleisli comparison functor takes its only object to this only algebra and sends $f\in M$ toIt is a bijection from the Kleisli arrows to the algebra endomorphisms, with inverse $h\mapsto (n\mapsto h(n+1))$. It preserves identities and composition by the comparison construction; directly, $K(\eta )=1$ and $K(g\star f)=K(g)K(f)$. Bijectivity on objects and arrows makes it an isomorphism of categories:

Let $M$ be the monoid of order-preserving functions $\mathbb{N}\to \mathbb{N}$, with $\mathbb{N}=\{0,1,\dots \}$, regarded as a one-object category. The endofunctor $T$ fixes that object and sends $f$ to $Tf(0)=0$, $Tf(n+1)=f(n)+1$. The displayed maps give its monad unit and multiplication. The multiplication is not injective, so this is not an idempotent monad. Its only algebra for a monad is $\mu$, because $a\eta =1$ forces $a(n+1)=n$ and monotonicity forces $a(0)=0$. Algebra endomorphisms are exactly the functions fixing zero. The Kleisli comparison functor sends $f$ to the function which is zero at zero and equals $f(n)$ at $n+1$; its inverse sends $h$ to $n\mapsto h(n+1)$. Thus the comparison is an isomorphism of categories even though the monad is not idempotent.