that the intersection $P(013200)\cap P(010065)$ is the single point $(\emptyset,1,21,21)$, which is contractible.

Now that we know the Möbius function of the intersections we can compute $\mu(A^{\sigma,\pi})$:

\[lem:muA\] $$\mu(A^{\sigma,\pi})=(-1)^{|\pi|-|\sigma|}\operatorname{\text{\rm NE}}(\sigma,\pi)+\sum_{S\in EZ^{\sigma,\pi}}(-1)^{|S|}.$$

We can split Equation into two parts: $$\label{eq:incex2}\mu(A^{\sigma,\pi})=\sum_{\eta\in \widehat{E}^{\sigma,\pi}}\mu(P(\eta)^o)+\sum_{\substack{S\subseteq \widehat{E}^{\sigma,\pi}\\|S|>1}}(-1)^{|S|-1}\mu(\bigcap_{\eta\in S}P(\eta)^o).$$ By Lemma \[lem:inter\] the second part of the right hand side of Equation equals $\displaystyle\sum_{S\in EZ^{\sigma,\pi}}(-1)^{|S|}$.

By the definition of $P(\eta)$, and the identity $\mu(A\times B)=\mu(A)\mu(B)$, we know $$\mu(P(\eta)^o)=\prod_{1\le i \le t} \mu(\hat{\eta}_i,\hat{\pi}_i).$$ We know that $[\hat{\eta}_i,\hat{\pi}_i]$ is always a chain, so by the definition of normality if $\eta$ is not normal there is some $i$ such that $|\hat{\eta}_i|\le|\hat{\pi}_i|-2$, so $\mu(\hat{\eta}_i,\hat{\pi}_i)=0$, which implies $\mu(P(\eta)^o)=0$. If $\eta$ is normal then $|\hat{\pi}_i|-|\hat{\eta}_i|=0$ or $1$, so $\mu(\hat{\eta}_i,\hat{\pi}_i)=1$ or $-1$, for all $i$. There are $|\pi|-|\sigma|$ parts $[\hat{\eta}_i,\hat{\pi}_i]$ with $\mu(\hat{\eta}_i,\hat{\pi}_i)=-1$, one for each zero in $\eta$, and the remaining have $\mu(\hat{\eta}_i,\hat{\pi}_i)=1$. Therefore, $\mu(P(\eta)^o)=(-1)^{|\pi|-|\sigma|}$ for each normal embedding, so the first term in the right hand side of Equation  equals $(-1)^{|\pi|-|\sigma|}\operatorname{\text{\rm NE}}(\sigma,\pi)$.

We now present our formula for the Möbius function that applies to any interval of permutations:

\[thm:main\] For any permutations $\sigma$ and $\pi$: $$\label{eq:main}\mu(\sigma,\pi)=(-1)^{|\pi|-|\sigma|}\operatorname{\text{\rm NE}}(\sigma,\pi)+\sum_{\lambda\in[\sigma,\pi)}\mu(\sigma,\lambda)\sum_{S\in EZ^{\lambda,\pi}}(-1)^{|S|}.$$

We take the poset $A^{\sigma,\pi}$ and for each $\lambda\in(\sigma,\pi)$ we retract $\widehat{E}^{\lambda,\pi}$ to a point 