$S^{2n}$, but every its proper subpolyhedron does.

Zaks’ series of examples satisfied the conclusion of Theorem \[Zaks\] except for being $(n-1)$-connected.

Let $K$ be the $n$-skeleton of the $(2n+2)$-simplex and let $\sigma$ be an $(n-1)$-simplex of $K$. By inspection, it is a face of at least three $n$-simplices; let $\tau$ be one of them. Let $b$ be the barycenter of $\sigma$, and let $B_i$ (resp. $D_i$) be the star of $b$ in the $i$th barycentric subdivision of $\tau$ (resp. of $\sigma$). Let $C_i$ be the closure of $|\partial B_i|\but |D_i|$; it is a codimension one ball properly embedded in $|\tau|$, with boundary sphere embedded in $|\sigma|$.

Given a positive integer $r$, let $P_r$ be the polyhedron obtained from $P_0\bydef|K|$ by shrinking each $C_i$ to a point $p_i$ for $i=1,\dots,r$. Then the quotient map $f_r{\colon}P_0\to P_r$ is finite-collapsible, and sends $|B_1|$ onto a collapsible polyhedron $X_r$. The quotient $P_r/X_r$ is homeomorphic to $P_0/|B_1|$ and therefore to $P_0$. Thus we obtain a finite-collapsible map $g_r{\colon}P_r\to P_0$.

The links of the points $p_i$ in $P_r$ are homeomorphic to each other and not homeomorphic to the link of any other point in $P_r$. Consequently, $P_0,P_1,P_2,\dots$ are pairwise non-homeomorphic. Since $P_0$ does not embed in 