Is the Möbius invariant of a poset, selfdual? The Möbius function $\mu_P$ of a finite bounded poset $P$ is defined recursively as follows:
$$\mu_P(x,x) = 1  ,  \forall x \in P$$ $$\mu_P(x,y) = - \sum_{x \le z < y} \mu_P(x,z) , \forall x < y \in P$$
The Möbius invariant $\mu(P)$ is $\mu_P(\hat{0},\hat{1})$, with $\hat{0}$ and $\hat{1}$, resp. the minimum and the maximum.
Let $P^{\star}$ be the dual poset of $P$ (i.e. with the inverse order).    
Question: Is it true that  $\mu(P) = \mu(P^{\star})$?  
 A: It is always true that $\mu_{P^*}(y,x)=\mu_P(x,y)$, where $P^*$ is the poset dual to $P$; this is immediate from the following

Lemma. If $x,y\in P$ and $x<y$, then $$\mu(x,y)=-\sum_{x<z\le y}\mu(z,y)\;.$$
Proof. Let $n$ be the length of the longest chain in $[x,y]$; the proof is by induction on $n$. If $n=2$, so that $y$ covers $x$, then $$\mu(x,y)=-\mu(x,x)=-1=-\mu(y,y)\;.$$ Now suppose that $n>2$, and the result holds for intervals in which the length $m$ of the longest chain satisfies $2\le m<n$. Then
$$\begin{align*}\mu(x,y)&=-\sum_{x\le z<y}\mu(x,z)\\
&=-\mu(x,x)-\sum_{x<z<y}\mu(x,z)\\
&=-1-\sum_{x<z<y}\mu(z,y)&\text{induction hyp.}\\
&=-\mu(y,y)-\sum_{x<z<y}\mu(z,y)\\
&=-\sum_{x<z\le y}\mu(z,y)\;,
\end{align*}$$
as desired. $\dashv$

Another approach is to observe that the incidence algebras of $P$ and $P^*$ contain the same functions, and $\zeta_{P^*}(y,x)=\zeta_P(x,y)$ for each interval $[x,y]$ in $P$. Since $\mu_P$ and $\mu^{P^*}$ are the inverses of $\zeta_P$ and $\zeta_{P^*}$, respectively, it follows that $\mu_{P^*}(y,x)=\mu_P(x,y)$.
