Let $R$ be a commutative ring and $M_R$ the set of monic polynomials on the indeterminate $q$. Let $M\subset M_R$ and denote by $M^*$ the multiplicative subset associated to $M$. Ordering all mentioned subsets by the divisibility relation gives directed sets. Define now a completion by $$R[q]^M=\varprojlim_{f\in M*} R[q]/(f)$$ Now, let $M'\subset M\subset M_R$ be another subset. Then $(M')^*$ is a directed subset of $M^*$, and denote its completion by $R[q]^{M'}$.
How does the identity map $\mathrm{id}_{R[q]}\colon R[q]\to R[q]$ induces a map $\rho_{M,M'}\colon R[q]^M\to R[q]^{M'}?$
I've been wondering about this for a while and I can't seem to define the map $\rho_{M,M'}$, the divisibility relation has to play some role, but I need some help. Thank you.
The $R[q]$-algebra $R[q]^{M}$ comes equipped with projection maps $\pi_{M,f} : R[q]^{M} \to R[q]/(f)$ for all $f \in M^{\ast}$. In particular we have $\pi_{M,f}$ for all $f \in (M')^{\ast}$ since $(M')^{\ast} \subset M^{\ast}$. Thus the universal property of $R[q]^{M'}$ gives us the desired $R[q]$-algebra homomorphism $\rho_{M,M'}$.