Given a finite poset $\mathcal P$, we say that a family $\mathcal F$ of subsets of $[n]$ is $\mathcal P$-saturated if $\mathcal F$ does not contain an induced copy of $\mathcal P$, but adding any other set to $\mathcal F$ creates an induced copy of $\mathcal P$. The saturation number of $\mathcal P$ is the size of the smallest $\mathcal P$-saturated family with ground set $[n]$. The saturation number for posets is known to exhibit a dichotomy: it is either bounded or it has at least $\sqrt n$ rate of growth. Determining which posets have bounded saturation number is a major open problem. In this paper we consider a `gluing' operation, formed from two finite posets $\mathcal P$ and $\mathcal Q$ by setting all elements of $\mathcal P$ to be below all elements of $\mathcal Q$. We show that (under some mild assumptions) this operation preserves bounded and unbounded saturation number. This is the first such `new from old' poset construction to be found. As an application, we show that for any poset $\mathcal P$ one may add at most 3 elements to $\mathcal P$ to obtain a poset whose saturation number growth is at most linear: this may be viewed as a step towards the other major open problem in the area, namely the conjecture that every finite poset has this growth at most linear. We also consider the poset equivalent of weak saturation for graphs: for each finite poset $\mathcal P$, we determine exactly the minimum size of a percolating family for $\mathcal P$.