Consider the finite Boolean lattice $B\_n$ of subsets of $[n]:=\lbrace 1,\dots,n\rbrace$ ordered by inclusion, let $1\leq j,k\leq n$ and consider the poset:
$$A\_{j,k}=\lbrace\emptyset\neq U\in B\_n\mid (1,\dots,j)\nsubseteq U, (n-k,\dots,n)\nsubseteq U \rbrace$$
I want to compute the geometric realization for every $j,k$. Using the ideas of ([Homotopy type of the geometric realization of a poset](https://mathoverflow.net/questions/448347/homotopy-type-of-the-geometric-realization-of-a-poset?noredirect=1#comment1160121_448347)) I was able to prove the following:


If $j\leq n-k-1$ then:
$$|A\_{j,k}|=|((B\_{j}\setminus\lbrace\hat{1}\rbrace)\times B\_{n-j-k-1}\times (B\_{k+1}\setminus\lbrace\hat{1}\rbrace))\setminus\lbrace\hat{0}\rbrace|=$$$$=|B\_{j}\setminus\lbrace\hat{1},\hat{0}\rbrace|\ast| B\_{n-j-k-1}\setminus\lbrace\hat{0}\rbrace|\ast |B\_{k+1}\setminus\lbrace\hat{1},\hat{0}\rbrace|=$$$$=\mathbb{S}^{j-2}\ast | B\_{n-j-k-1}\setminus\lbrace\hat{0}\rbrace|\ast \mathbb{S}^{k-1}=\mathbb{S}^{j+k-2}\ast | B\_{n-j-k-1}\setminus\lbrace\hat{0}\rbrace|$$
Then, if $n=j+k+1~|A\_{j,k}|=\mathbb{S}^{j+k-2}=\mathbb{S}^{n-3}$ and if $n\neq j+k+1~|A\_{j,k}|$ is contractible.


But I don't know how we can prove it for $j\geq n-k$, since in this case the two sets $(1,\dots,j)$ and $(n-k,\dots,n)$ intersect, so I can't split $A\_{j,k}$ into a direct product.


Any hint or help will be thanked.


Edit: Using the hints of @ChristopheLeuridan and @DavidWhite I did the following for the case $j\geq n-k$:
First of all, we can decompose $A\_{j,k}=A\_{j,k}^1\cup A\_{j,k}^2$ where:
$$A\_{j,k}^1=\lbrace\emptyset\neq U\in B\_n\mid (n-k,\dots,j)\nsubseteq U\rbrace$$
$$A\_{j,k}^2=\lbrace\emptyset\neq U\in B\_n\mid (n-k,\dots,j)\subseteq U\text{ and }(1,\dots,n-k-1)\nsubseteq U,(j+1,\dots,n)\nsubseteq U\rbrace$$
We can compute the geometric realization of those two subposets:
$$|A\_{j,k}^1|=|((B\_{j-n-k+1}\setminus\lbrace\hat{1}\rbrace)\times B\_{2n-j+k-1})\setminus\lbrace\hat{0}\rbrace|=$$
$$=|B\_{j-n-k+1}\setminus\lbrace\hat{0},\hat{1}\rbrace|\ast|B\_{2n-j+k-1}\setminus\lbrace\hat{0}\rbrace|=\mathbb{S}^{j-n+k-3}\ast|B\_{2n-j+k-1}\setminus\lbrace\hat{0}\rbrace|$$
Hence, if $2n-j+k-1=0~|A\_{j,k}^1|=\mathbb{S}^{j-n+k-3}=\mathbb{S}^{n+2k-4}$ and if $2n-j+k-1\neq0~|A\_{j,k}^1|$ is contractible, but it is easy to see that we can't have $2n-j+k-1=0$. On the other hand:
$$|A\_{j,k}^2|=|((B\_{n-k-1}\setminus\lbrace\hat{1}\rbrace)\times(B\_{n-j}\setminus\lbrace\hat{1}\rbrace))\setminus\lbrace \hat{0}\rbrace|=$$
$$=|B\_{n-k-1}\setminus\lbrace\hat{0},\hat{1}\rbrace|\ast|B\_{n-j}\setminus\lbrace\hat{0},\hat{1}\rbrace|=\mathbb{S}^{n-k-3}\ast\mathbb{S}^{n-j-2}=\mathbb{S}^{2n-k-j-4}$$
Hence, since those two posets are disjoint we have:
$$|A\_{j,k}|=|A\_{j,k}^1|\cup|A\_{j,k}^2|=\mathbb{S}^{2n-k-j-4}$$



Geometric realization of a poset
Put $P\_0=\{1,\dotsc,n-k-1\}$ and $P\_1=\{n-k,\dotsc,j-1\}$ and $P\_2=\{j,\dotsc,n\}$, so $[n]=P\_0\amalg P\_1\amalg P\_2$ with each $P\_i$ nonempty. Any subset $U$ can be decomposed as $\coprod\_{i=0}^2U\_i$ with $U\_i\subseteq P\_i$. You are looking at the space
$$ A = \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset, U\_0\cup U\_1\neq P\_0\cup P\_1, U\_1\cup U\_2\neq P\_1\cup P\_2\}. $$
This can be written as $B\cup C$, where
\begin{align\*}
 B &= \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset,U\_1\neq P\_1\} \\
 C &= \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset,U\_0\neq P\_0,U\_2\neq P\_2\} \\
 D &= B\cap C = \{(U\_0,U\_1,U\_2): U\_0\cup U\_1\cup U\_2\neq\emptyset,U\_0\neq P\_0,U\_1\neq P\_1,U\_2\neq P\_2\}
\end{align\*}
We can define $f\_0,f\_1,f\_2\colon B\to B$ by $f\_0=\text{id}$ and $f\_1(U\_0,U\_1,U\_2)=(P\_0,U\_1,P\_2)$ and $f\_2(U\_0,U\_1,U\_2)=(P\_0,\emptyset,P\_2)$. We have $f\_0\leq f\_1\geq f\_2$, so the identity map $|f\_0|\colon|B|\to|B|$ is homotopic to the constant map $|f\_2|$, so $B$ is contractible. Similarly, we can define $g\_0,g\_1,g\_2\colon C\to C$ by $g\_0=\text{id}$ and $g\_1(U\_0,U\_1,U\_2)=(U\_0,P\_1,U\_2)$ and $g\_2(U\_0,U\_1,U\_2)=(\emptyset,P\_1,\emptyset)$, and use these to see that $|C|$ is contractible. We also have a pushout square of subcomplex inclusions
$\require{AMScd}$
\begin{CD}
|D| @>>> |B|\simeq \* \\
@VVV @VVV\\
 |C|\simeq \* @>>> |A|
\end{CD}
which gives $|A|\simeq\Sigma|D|$. Now put $D\_i=\{U\_i : \emptyset\subset U\_i\subset P\_i\}$, so $|D\_i|\simeq S^{|P\_i|-2}$. It is easy to see that $|D|$ is the same as the join $|D\_0|\*|D\_1|\*|D\_2|$, and $S^r\*S^s\simeq S^{r+s+1}$, so $|D|\simeq S^{|P\_0|+|P\_1|+|P\_2|-4}=S^{n-4}$ and $|A|\simeq\Sigma S^{n-4}=S^{n-3}$.


