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Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.

Approach 1.

Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.

Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.

This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands on $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.

Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.

A natural question here is whether a characterization is known of those general (non-distributive) lattices which do not possess nontrivial dense (neither codense) elements.

Note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.

In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.

Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...

Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.

Approach 1.

Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.

Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.

This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands on $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.

Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.

A natural question here is whether a characterization is known of those general (non-distributive) lattices which do not possess nontrivial dense (neither codense) elements.

Note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.

In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.

Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...

Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.

Approach 1.

Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.

Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.

This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands in $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.

Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.

A natural question here is whether a characterization is known of those general (non-distributive) lattices which do not possess nontrivial dense (neither codense) elements.

Note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.

In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.

Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...

Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.

Approach 1.

Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.

Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.

This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands on $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.

Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.

A natural question here is whether a characterization is known of those general (non-distributive) lattices which do not possess nontrivial dense (neither codense) elements.

Note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.

In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.

Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...

Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.

Approach 1.

Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.

Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.

This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands in $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.

Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.

However note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.

In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.

Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...

Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.

Approach 1.

Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.

Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.

This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands in $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.

Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.

A natural question here is whether a characterization is known of those general (non-distributive) lattices which do not possess nontrivial dense (neither codense) elements.

Note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.

In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.

Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...