I have a question concerning proposition 10.14 from Steve Awodey's lecture notes (PDF) concerning the relationship between algebras for endofunctors and algebras for monads. Specifically, the proposition states that the following are equivalent for an endofunctor $P : \mathbf{S} \to \mathbf{S}$:
- There is a categorical equivalence $\text{Alg}(P) \cong \text{Alg}(T)$ for a monad $T : \mathbf{S} \to \mathbf{S}$
- The forgetful functor $U : \text{Alg}(P) \to \mathbf{S}$ has a left adjoint
- For each $A \in \mathbf{S}$, the endofunctor $P_A : X \mapsto A + P(X)$ has an initial algebra.
Implicit in the theorem (or detailed in the proof, I believe) is that:
- The monad $T = UF$ if $F$ is the left adjoint.
- The action of $T$ takes an object $A$ to the initial algebra of the endofunctor $P_A$ mentioned in the third point above, which is the "free algebra on $A$" ("free monads are initial algebras")
Unfortunately I have trouble following this theorem through for the example of monoids. Specifically, let $P(X) = 1 + X^2$ (work over $\mathbf{S} = \mathbf{Set}$ for simplicity). An algebra for $P$ is a monoid structure on a set. The associated monad $T$ should be (I think), the "list" or "free monoid" functor $TA = A^*$, the monad whose algebras are traditionally recognized as monoids. I have two questions:
- I do not see how $T$ defined by $TA = A^*$ is the initial algebra for $P_A$, which maps $A \mapsto 1 + P(X) = A + (1 + X^2)$, i.e. how $A^*$ the least fixed point/solution to $X = A + (1 + X^2)$. It seems instead that $A^*$ is the solution to $X = 1 + A \times X$.
- It is my impression that $\text{Alg}(P)$ contains sets with monoid structure that are, in particular, not even monoids, whereas $\text{Alg}(T)$ imposes more restrictions (algebra arrows must be compatible with the unit and multiplication of the monoid) that actually restrict such maps to be proper monoids (e.g., associative, and that the unit acts like a unit) So how can we have the equivalence $\text{Alg}(P) \cong \text{Alg}(T)$?