We often use the following obvious fact.

\[subposet\] Let $P$ be a simplicial poset and $A=K[x_\sigma: \sigma \in P \setminus \{\hat 0\}]$. If $Q$ is an order ideal of $P$ then $A/(I_P+(x_\sigma: \sigma \not \in Q))$ is isomorphic to $K[Q]$ as a ring.

Let $P$ be a simplicial cell $(d-1)$-ball. Then each rank $d-1$ element of $P$ is covered by at most two elements. The *boundary $\partial P$ of $P$* is the order ideal $$\partial P=\langle \sigma \in P: {\ensuremath{\mathrm{rank}}\hspace{1pt}}\sigma=d-1,\ \sigma \mbox{ is covered by exactly one element in }P\rangle.$$ Since $\partial P$ is the face poset of the boundary cell complex of $\Gamma(P)$, $\partial P$ is a simplicial cell $(d-2)$-sphere. Moreover, it is known that if ${\mathbf{h}}=(h_0,h_1,\dots,h_d)$ is the $h$-vector of $P$ then $\partial {\mathbf{h}}=(\partial h_0,\partial h_1,\dots,\partial h_{d-1})$ is the $h$-vector of $\partial P$. See [@Ko Section 3]. On $h$-vectors of simplicial cell balls, the following result was proved in [@Ko Theorem 5].

\[kolins\] Let $P$ be a simplicial cell $(d-1)$-ball and ${\mathbf{h}}=h(P)=(h_0,h_1,\dots,h_d)$. Then

-   $h_0=1,$ $h_d=0$ and $h_k \geq 0$ for $k=1,2,\dots,d-1$.

-   $\partial h_k \geq 0$ for $k=0,1,\dots,d-1$.

-   If $d$ is odd and $\partial h_n=0$ for some $1 \leq n \leq d-2$ then $\sum_{k=0}^d h_k$ is 