Geometric realization of a poset

Question

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) 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: $$|A_{j,k}|=|A_{j,k}^1|\cup|A_{j,k}^2|=\mathbb{S}^{2n-k-j-4}$$

Answer 1

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 $|A|\simeq\Sigma S^{n-4}=S^{n-3}$.