Q = (P \cup Q) /\! \sim_f.$$ Then $P \cup_f Q$ is again a simplicial poset. We write $P \cup _I Q$ (or $P {{}_I \cup_J} Q$) for this simplicial poset $P \cup_f Q$ when the isomorphism $f$ is clear.

\[3.1\] Let $P$ and $Q$ be $(d-1)$-dimensional simplicial posets and let $I \subset P$ and $J \subset Q$ be order ideals with $I \cong J$. If $\dim I=d-2$ then $$h_k(P\cup_I Q)=h_k(P) + h_k(Q) -h_k(I) + h_{k-1}(I)$$ for $k=0,1,\dots,d$, where $h_{-1}(I)=h_d(I)=0$.

The desired formula follows from $f_k(P \cup_I Q)=f_k(P)+f_k(Q)-f_k(I)$ by straightforward computations.

For a positive integer $i$, we write $[i]=\{1,2,\dots,i\}$. For integers $1 \leq k \leq d$, let $$\Delta_d(k)=
\big\{F \subset [d]: F \not \supset [k]\big\}.$$ By defining a partial order on $\Delta_d(k)$ by inclusion, we regard $\Delta_d(k)$ as a simplicial poset of dimension $d-2$. In other words, $\Delta_d(k)$ is the (abstract) simplicial complex generated by $\{ [d] \setminus \{i\}: i=1,2,\dots,k\}$. Let ${\mathbf{e}}_0,{\mathbf{e}}_1,\dots,{\mathbf{e}}_d \in { \ensuremath{\mathbb{Z}}}^{d+1}$ be the unit vectors of ${ \ensuremath{\mathbb{Z}}}^{d+1}$. Thus ${\mathbf{e}}_i$ is the vector such that $(i+1)$-th entry of ${\mathbf{e}}_i$ is $1$ and all other entries of ${\mathbf{e}}_i$ are $0$. It is easy to see that $h_i(\Delta_d(k))=1$ for $i < k$ and $h_i(\Delta_d(k))=0$ for $i \geq k$. 