The free unital ring functor is left adjoint to the underlying-set functor

Send a word $[x_1,\dots ,x_n]$ to $u(x_1)\cdots u(x_n)$ in $R$, and send the empty word to $1_R$. The generator $x$ is represented by the basis word $[x]$.

Apply [L2] over $\mathbb{Z}$ to extend this word-evaluation function uniquely to a $\mathbb{Z}$-linear map $\bar{u}:{\mathbb{Z}}^{(X^{\ast })}\to R$.

Expanding convolution as a finite sum shows $\bar{u}(\alpha \beta )=\bar{u}(\alpha )\bar{u}(\beta )$, and the empty-word basis vector maps to $1_R$. Thus $\bar{u}$ is a unit-preserving ring homomorphism, including when $R$ is the zero ring.

Any ring homomorphism extending $u$ must send every word basis vector to the corresponding product and is additive, so it equals $\bar{u}$ by uniqueness in [L2].

Hence the generator inclusion is a universal arrow from every set $X$ to the underlying-set functor on rings. By [L3] these universal arrows assemble into the free-ring functor and the asserted adjunction; for $X=\mathrm{\varnothing }$ the free ring is $\mathbb{Z}$.