Let $W_0$ be the monad in $\textsf{Set}$ defined by forgetful functor $\textsf{Mon}$ to $\textsf{Set}$. Show that a $W_0$-algebra is a set $M$ with a string $v_0,v_1,\cdots$ of $n$-ary operations $v_n$, where $v_0:*\to M$ is the unit of monoid $M$ and $v_n$ is $n$-fold product.
I am solving problem in chapter 5 of MacLane's categories regarding to monad. My question is what unit of monoid $v_0:*\to M$ means. So far, I convince that $*$ is identity of free monoid $FM$, where $F$ is left adjoint of forgetful functor so $FM$ means free monoid of $M$ and I wonder "unit of monoid" merely implies what I mentioned or has further meaning.
In addition, I figured out if a set $M$ with a string $v_0,v_1,\cdots$ satisfies (1) $v_1=1$ (identity function) and (2) $v_k(v_{n_1}\times \cdots v_{n_k})=v_{n_1+\cdots +n_k}$ (this needs for associative law of $W_0$-algebra.), it becomes $W_0$-algebra whatever the image of $v_0:*\to M$ is.
Is the second paragraph can be an answer of problem???
The point is that, for whatever (equationally defined) algebraic structures and their corresponding free-forgetful adjunction, the algebras of its monad are just the original algebras.
So, in our case, it will turn out that $W_0$-algebras can naturally be identified with monoids.
Specifically, for a $W_0$-algebra $\langle M;\, v_0,v_1,v_2,\dots\rangle$ one associates the monoid $\langle M;\, v_0,v_2\rangle$ - where the 0-ary [i.e. constant] operation $v_0$ picks the unit of the monoid.
(Note that (2) also has to hold in cases $n_i=0$ or $k=0$, which will imply that the element picked by $v_0$ must be identity for the operation $v_2$.)
For the other direction, as you observed, we have to take $v_n$ to be the $n$-fold monoid operation (which is uniquely determined because of associativity).