Consider a set of $n$ elements $S=\lbrace 1,\dots,n\rbrace$ and $\mathcal{P}(S)$ to be the power set of $S$, which is a well-defined poset with respect to the inclusions. Now consider $\emptyset\neq T\varsubsetneq S$ and define $A=\lbrace \emptyset\neq U\in\mathcal{P}(S)\mid T\nsubseteq U\rbrace$. I want to compute the homotopy type of the geometric realization $\left|A\right|$.


My intuition tells me that $\left|A\right|\simeq\mathbb{S}^{\left|T\right|-2}$, however I don't know if this is right and how to prove it.


I know some facts such that if the poset has a maximal element or a minimal element then the geometric realization is contractible. Also, for some small cases I am able to do it by hand, however I don't know how to do it in general and whether there is some ''general'' procedure to study this problem.


Any help or references about this problem will be appreciated.



Homotopy type of the geometric realization of a poset
First recall that geometric realisation of posets preserves products: the projections $P\xleftarrow{p}P\times Q\xrightarrow{q}Q$ give a map $(|p|,|q|)\colon |P\times Q|\to|P|\times|Q|$, and it is a standard fact that this is a homeomorphism.


Next, if $f\_0,f\_1\colon P\to Q$ are morphisms of posets and $f\_0(x)\leq f\_1(x)$ for all $x$ then we can define a poset morphism $h\colon \{0,1\}\times P\to Q$ by $h(i,x)=f\_i(x)$, and this gives a map $|h|\colon [0,1]\times|P|\to|Q|$ which is a homotopy between $|f\_0|$ and $|f\_1|$.


Now put $A=\{U\subseteq S\;:\;U\neq\emptyset, U\not\supseteq T\}$ as in the question. We can define $f\_0,f\_1,f\_2\colon A\to A$ by $f\_0(U)=U$ and $f\_1(U)=U\cup T^c$ and $f\_2(U)=T^c$. Then $f\_0\leq f\_1\geq f\_2$, so the identity is homotopic to the constant map $|f\_2|$, so $|A|$ is contractible.


Now suppose instead we consider the poset $B=\{U\subseteq S\;:\;T\cap U\neq\emptyset,U\not\supseteq T\}\subset A$, and the poset $C=\{U\;:\;\emptyset\subset U\subset T\}$. We have an inclusion $i\colon C\to B$ and a retraction $r\colon B\to C$ given by $r(U)=T\cap U$. These have $ri=1$ and $ir\leq 1$ so $|i|$ and $|r|$ are mutually inverse homotopy equivalences.
Alternatively, we can say that $B\simeq C\times D$, where $D$ is the poset of all subsets of $T^c$. This gives $|B|\simeq|C|\times|D|$, where $|D|$ is homeomorphic to $[0,1]^{|T^c|}$ and in particular is contractible.


Now put $M=\{m:T\to\mathbb{R}\;:\;\sum\_{t\in T}m(t)=0\}$ (which is a vector space of dimension $|T|-1$ with an obvious inner product). For $U\in C$ define $f(U)=\chi\_U-|U||T|^{-1}\in M$. Extend this linearly over simplices to get a map $f\colon |C|\to M$. One can check that this is nowhere zero, so we can define $f\_1(x)=f(x)/\|f(x)\|$, and this gives a homeomorphism $|C|\to S(M)\simeq S^{|T|-2}$.


