[proofplan]
We partition the Möbius cancellation sum over $L$ according to the value of $x\vee a$. For each $y\ge a$, we introduce the fiber sum over all $x$ with $x\vee a=y$, and we show that every fiber sum with $y<\hat{1}$ vanishes. The key point is that the cumulative sum of these fibers below a fixed $y$ is exactly the usual Möbius cancellation sum over the interval $[\hat{0},y]$. After all proper fibers vanish, the total cancellation over $[\hat{0},\hat{1}]$ leaves precisely the desired fiber over $\hat{1}$.
[/proofplan]
[step:Define the fiber sums for the join map with $a$]
Let
\begin{align*}
U=\{y\in L:a\le y\}.
\end{align*}
For each $y\in U$, define the fiber sum $S(y)\in \mathbb{Z}$ by
\begin{align*}
S(y)=\sum_{\substack{x\in L, x\vee a=y}}\mu_L(\hat{0},x).
\end{align*}
This is a finite sum because $L$ is finite.
We first record the relation between the fibers of the map $x\mapsto x\vee a$ and intervals in $L$. If $y\in U$, then
\begin{align*}
\{x\in L:x\vee a\le y\}=\{x\in L:x\le y\}.
\end{align*}
Indeed, if $x\vee a\le y$, then $x\le x\vee a\le y$. Conversely, if $x\le y$ and $a\le y$, then $y$ is an upper bound for $\{x,a\}$, so the least upper bound $x\vee a$ satisfies $x\vee a\le y$.
Therefore, for every $y\in U$,
\begin{align*}
\sum_{\substack{z\in U, z\le y}}S(z)=\sum_{\substack{x\in L, x\le y}}\mu_L(\hat{0},x).
\end{align*}
The equality follows by grouping each element $x\le y$ according to the value $z=x\vee a$, which lies in $U$ and satisfies $z\le y$.
[/step]
[step:Show that every proper fiber sum below $\hat{1}$ vanishes]
We prove that
\begin{align*}
S(y)=0
\end{align*}
for every $y\in U$ with $y<\hat{1}$.
Fix such a $y$. Since $a\le y$ and $\hat{0}<a$, we have $\hat{0}<y$. By the recursive cancellation property of the Möbius function on a nontrivial interval, as in [citetheorem:8093],
\begin{align*}
\sum_{\substack{x\in L, x\le y}}\mu_L(\hat{0},x)=0.
\end{align*}
Using the cumulative fiber identity from the previous step gives
\begin{align*}
\sum_{\substack{z\in U, z\le y}}S(z)=0.
\end{align*}
We now use induction over the finite poset $U\setminus\{\hat{1}\}$, ordered by the order inherited from $L$. Suppose $y\in U$ with $y<\hat{1}$, and assume that $S(z)=0$ for every $z\in U$ with $z<y$. Then
\begin{align*}
0=\sum_{\substack{z\in U, z\le y}}S(z)=S(y)+\sum_{\substack{z\in U, z<y}}S(z)=S(y).
\end{align*}
Thus $S(y)=0$. Since $U\setminus\{\hat{1}\}$ is finite, induction proves the assertion for every $y\in U$ with $y<\hat{1}$.
[guided]
We want to prove that each proper fiber
\begin{align*}
\{x\in L:x\vee a=y\}
\end{align*}
has total Möbius weight zero when $a\le y<\hat{1}$. The fiber itself is not usually an interval, so we do not apply Möbius cancellation directly to that fiber. Instead, we apply cancellation to the cumulative union of all fibers below $y$.
Fix $y\in U$ with $y<\hat{1}$. Since $a\le y$ and $\hat{0}<a$, transitivity gives $\hat{0}<y$. The interval $[\hat{0},y]$ is therefore nontrivial. The [recursive formula for the Möbius function](/theorems/8093), stated in [citetheorem:8093], gives the cancellation identity
\begin{align*}
\sum_{\substack{x\in L, \hat{0}\le x\le y}}\mu_L(\hat{0},x)=0.
\end{align*}
Because $\hat{0}$ is the least element of $L$, the condition $\hat{0}\le x$ is automatic for $x\in L$, so this is
\begin{align*}
\sum_{\substack{x\in L, x\le y}}\mu_L(\hat{0},x)=0.
\end{align*}
Now we translate this interval sum into fiber sums. From the previous step, for this same $y$ we have
\begin{align*}
\sum_{\substack{z\in U, z\le y}}S(z)=\sum_{\substack{x\in L, x\le y}}\mu_L(\hat{0},x).
\end{align*}
Combining the two equalities gives
\begin{align*}
\sum_{\substack{z\in U, z\le y}}S(z)=0.
\end{align*}
This identity gives the cumulative sum of all fiber weights below $y$, not immediately the single fiber weight $S(y)$. To isolate $S(y)$, use induction in the finite poset $U\setminus\{\hat{1}\}$. Assume that for the chosen $y$, every smaller element $z\in U$ with $z<y$ already satisfies $S(z)=0$. Then the cumulative identity becomes
\begin{align*}
0=\sum_{\substack{z\in U, z\le y}}S(z)=S(y)+\sum_{\substack{z\in U, z<y}}S(z)=S(y).
\end{align*}
Thus $S(y)=0$. The base cases are included in the same argument: if no element of $U$ lies strictly below $y$, then the lower sum is empty and the displayed equality gives $S(y)=0$ directly. Since $U\setminus\{\hat{1}\}$ is finite, this induction proves that every proper fiber sum below $\hat{1}$ vanishes.
[/guided]
[/step]
[step:Partition the total Möbius cancellation and keep only the top fiber]
Since $\hat{0}<a\le\hat{1}$, we have $\hat{0}<\hat{1}$. Applying the same Möbius cancellation identity to the nontrivial interval $[\hat{0},\hat{1}]$ gives
\begin{align*}
\sum_{x\in L}\mu_L(\hat{0},x)=0.
\end{align*}
Every element $x\in L$ has $x\vee a\in U$, so partitioning this finite sum by the value of $x\vee a$ yields
\begin{align*}
0=\sum_{x\in L}\mu_L(\hat{0},x)=\sum_{y\in U}S(y).
\end{align*}
By the previous step, $S(y)=0$ for every $y\in U$ with $y<\hat{1}$. Therefore
\begin{align*}
0=S(\hat{1})=\sum_{\substack{x\in L, x\vee a=\hat{1}}}\mu_L(\hat{0},x).
\end{align*}
This is the desired identity.
[/step]
