result with a different condition on the poset fibration that generalises the argument in the proof of Theorem \[thm:main\], this result can be seen as an application of [@Wal81 Corollary 3.2]. We let $f^*$ and $f^{-1}$ denote the image and preimage of $f$, respectively.

\[prop:mob2.5\] Let $f:P\rightarrow Q$ be a surjective poset map such that $f^*(P_{<p})=Q_{<q}$, for any $q\in Q$ and $p\in f^{-1}(q)$. Then $$\mu(Q)=\mu(P)+\sum_{q\in Q}\mu(Q_{<q})\mu(f^{-1}(Q_{\ge q})).$$

We begin with $P$ and for each $q\in Q$ we retract $f^{-1}(q)$ to a single point and observe the effect this retraction has on the Möbius function of $P$. We do this inductively from the bottom to the top, so when considering $q\in Q$ we assume all points in $P_{<p}$, for all $p\in f^{-1}(q)$, have been retracted.

To calculate the effect the retraction has on the Möbius function of the poset we consider $$W(q):=\{p\in P\,|\, p<\lambda\text{ or }p\ge\lambda\text{ for some }\lambda\in f^{-1}(q)\}.$$ When we retract $f^{-1}(q)$ to a point we retract $W(q)$ to a contractible poset, which implies the change to the Möbius function is $-\mu(W(q))$. We can rewrite $W(q)$ in the following way: $$\begin{aligned}
W(q)&=\bigcup_{p\in f^{-1}(q)}P_{<p}\star P_{\ge p}=\bigcup_{p\in f^{-1}(q)}Q_{<q}\star P_{\ge p}\\&=Q_{<q}\star\bigcup_{p\in f^{-1}(q)} P_{\ge p}=Q_{<q}\star f^{-1}(Q_{\ge q}).\end{aligned}$$ We can replace $P_{<p}$ with $Q_{<q}$ 