The only reaction to [this question](https://math.stackexchange.com/q/1867404/214353) on math.SE was 21 views in 4 days, so I decided to repost it here. I am not changing anything.


There was an interesting question on MO which OP removed by some reason. Here is a (more or less) equivalent form.


Take a finite cartesian product of finite linear orders, and remove top and bottom. What is the homotopy type of the obtained poset?


When all linear orders have two elements, this is a sphere. In that question, it was conjectured that if at least one of them has more than two elements, then it is contractible. I think in fact from $3\times2\times2\times2$ one gets a (thickened) 2-sphere but I am not sure.


In general, is it known which homotopy types may occur when one removes top and bottom from a finite lattice?


**Later**:


In the course of answers, the following additional question became relevant (I think).


Call an element of a bounded lattice dense if it has nonbottom meet with every nonbottom element. Top is obviously such; call a dense element nontrivial if it is not top.


It is known that a finite distributive lattice (more generally a not necessarily finite [pseudocomplemented](https://en.wikipedia.org/wiki/Pseudocomplement) lattice) does not have nontrivial dense elements if and only if it is a Boolean algebra.


Is a generalization of this known for arbitrary lattices? That is, which general lattices do not have any nontrivial dense elements?


**Still later:**


I've posted a question about the last one, [Lattices without nontrivial dense elements](https://mathoverflow.net/q/245366/41291); it was answered by [Tri](https://mathoverflow.net/users/51389/tri). So seems like correct type of lattices to consider in this context are the [geometric lattices](https://en.wikipedia.org/wiki/Geometric_lattice).



Homotopy type of some lattices with top and bottom removed
Let $C(P)$ be the poset obtained by removing top and bottom from $P$ (where $P$ is a poset having a top and a bottom, not equal). Then $C(P\times Q)$ is homotopy equivalent to $\Sigma(C(P)\ast C(Q))$, the suspension of the join. Thus if one of the linear orders in your product has at least three elements then $C$ of the product will indeed be contractible, since the join of $X$ with a contractible space is always contractible. Note that the join of $X$ with the empty space is $X$.


