Prove forgetful functor U: Monoid -> Set has a left adjoint.

I has been given the definition of having a left adjoint:

A functor $U:\mathbf{C}\to\mathbf{D}$ has a left adjoint if for all $\mathbf{X} \in \mathbf{D}$, there exist $\mathbf{FX} \in \mathbf{C}$ and unit $\mathbf{ηx}: \mathbf{X}\to \mathbf{UFX}$ in $\mathbf{D}$ such that for all $\mathbf{A} \in \mathbf{C}$ and for all $\mathbf{f}: \mathbf{X}\to \mathbf{UA}$, there exists a unique map $\mathbf{g}: \mathbf{FX}\to \mathbf{A}$. such that the diagram commutes.

C is the category Monoid of monoids, D is the category Set and U is the forgetful functor.

$U:\mathbf{Mon}\to\mathbf{Set}$

What will be $\mathbf{FX}$, $\mathbf{ηx}$ and $\mathbf{g}$ in this case?

• Hint: left adjoint to forgetful functor = free functor. So, what is a free monoid? – Ennar Feb 27 '17 at 21:09
• so FX is the free monoid (set of lists of elements of the set X with monoid operation concatenation). What will ηx be? – F. Zhao Feb 27 '17 at 21:42
• Well, what is the most natural list to assign to some element $x\in X$? – Ennar Feb 27 '17 at 21:44
• Is it ηx: x -> {x}? – F. Zhao Feb 27 '17 at 21:47
• Precisely. Now just prove everything checks out. – Ennar Feb 27 '17 at 21:48