Christopher Upshaw wrote:

>I don't think \\(\mathbf{Grp}[L(m),g] \cong \mathbf{Mon}[m,R(g)]\\) is true with \\(\mathbf{L} = \mathbf{T}\circ\mathbf{S}\\).

>Let's look at \\(m = (bool, or), g = L(m)\\).

>Given \\(f : m\to g\\) \\[f(T) = f (TT) =f(T)f(T)\\] but the only element of g with that property is \\(\epsilon\\). So there is exactly one morephism on the rhs.

>But on the lhs, there are at least two, \\( x\mapsto \epsilon\\) and \\(x \mapsto x\\).

I don't understand your reasoning here. In which category does \\(f\\) live in? And why are we using \\(T\\)?

>I don't think \\(\mathbf{Grp}[L(m),g] \cong \mathbf{Mon}[m,R(g)]\\) is true with \\(\mathbf{L} = \mathbf{T}\circ\mathbf{S}\\).

>Let's look at \\(m = (bool, or), g = L(m)\\).

>Given \\(f : m\to g\\) \\[f(T) = f (TT) =f(T)f(T)\\] but the only element of g with that property is \\(\epsilon\\). So there is exactly one morephism on the rhs.

>But on the lhs, there are at least two, \\( x\mapsto \epsilon\\) and \\(x \mapsto x\\).

I don't understand your reasoning here. In which category does \\(f\\) live in? And why are we using \\(T\\)?