Return to Answer

The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.

Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.

In fact, the corresponding statement is true for monads on any category, not just $\mathrm{SET}$.

The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.

Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.

In fact, the corresponding statement is true for monads on any category, not just $\mathrm{Set}$.

The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.

Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.

The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.

Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.

In fact, the corresponding statement is true for monads on any category, not just $\mathrm{SET}$.

The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.

The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.

Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.