[proofplan] Define an auxiliary function $F:\{(u,v)\in P\times P:u\le v\}\to\mathbb Z$ by setting $F(u,u)=1$ and, for $u<v$, $F(u,v)=\widetilde{\chi}\bigl(\Delta((u,v))\bigr)$. We expand $F(u,v)$ as an alternating sum over chains strictly between $u$ and $v$, with the reduced Euler characteristic contributing the empty-chain term. The main point is that this alternating chain sum satisfies the same interval recursion as the Möbius function. Since the recursion with the diagonal condition determines the Möbius function uniquely on a finite interval, this gives $F(x,y)=\mu_P(x,y)$. [/proofplan]

[step:Expand the reduced Euler characteristic as an alternating chain sum] Define the auxiliary function $F:\{(u,v)\in P\times P:u\le v\}\to \mathbb Z$ as follows: for $u\le v$, set $F(u,u)=1$ when $u=v$, and set $F(u,v)=\widetilde{\chi}(\Delta((u,v)))$ when $u<v$, where $(u,v)=\{z\in P:u<z<v\}$ and $\Delta((u,v))$ is the order complex of this open interval. Fix $u<v$. For each integer $k\ge 1$, let $c_k(u,v)$ denote the number of chains $u<z_1<\cdots<z_k<v$ in $P$ with exactly $k$ elements strictly between $u$ and $v$. Since $P$ is finite, only finitely many of the integers $c_k(u,v)$ are nonzero. A chain with $k$ internal elements is a simplex of $\Delta((u,v))$ of dimension $k-1$. Therefore the reduced Euler characteristic is $F(u,v)=-1+\sum_{k\ge 1}(-1)^{k-1}c_k(u,v)$. The term $-1$ is the reduced Euler characteristic contribution of the empty simplex convention. In particular, if there is no element $z\in P$ satisfying $u<z<v$, then $(u,v)=\varnothing$, all $c_k(u,v)$ vanish, and $F(u,v)=-1$, as required by the reduced convention. [/step]

[step:Verify the interval recursion by grouping chains according to their last element]We prove that for every $u<v$ in $P$, \begin{align*} F(u,v)=-\sum_{u\le z<v}F(u,z). \end{align*} Equivalently, \begin{align*} \sum_{u\le z\le v}F(u,z)=0. \end{align*} Using $F(u,u)=1$ and the chain expansion for each $u<z<v$, we have \begin{align*} -\sum_{u\le z<v}F(u,z)=-1-\sum_{u<z<v}F(u,z). \end{align*} For each $z$ with $u<z<v$, the expansion of $F(u,z)$ is \begin{align*} F(u,z)=-1+\sum_{m\ge 1}(-1)^{m-1}c_m(u,z). \end{align*} Substituting this into the preceding expression gives \begin{align*} -\sum_{u\le z<v}F(u,z) = -1+\#\{z\in P:u<z<v\} -\sum_{u<z<v}\sum_{m\ge 1}(-1)^{m-1}c_m(u,z). \end{align*} The middle term counts the chains $u<z_1<v$ with exactly one internal element, so it is $c_1(u,v)$. For $k\ge 2$, every chain \begin{align*} u<z_1<\cdots<z_k<v \end{align*} has a unique last internal element $z_k$, and its prefix \begin{align*} u<z_1<\cdots<z_{k-1}<z_k \end{align*} is counted in $c_{k-1}(u,z_k)$. Conversely, every chain counted by some $c_{k-1}(u,z)$ with $u<z<v$ extends uniquely to a chain \begin{align*} u<z_1<\cdots<z_{k-1}<z<v. \end{align*} Thus, for each $k\ge 2$, the total contribution of chains with $k$ internal elements is $(-1)^{k-1}c_k(u,v)$. Hence \begin{align*}-\sum_{u\le z<v}F(u,z)=-1+\sum_{k\ge 1}(-1)^{k-1}c_k(u,v)=F(u,v).\end{align*}[/step]

[guided]Recall the auxiliary function $F:\{(a,b)\in P\times P:a\le b\}\to\mathbb Z$: it is defined by $F(a,a)=1$, and for $a<b$ by $F(a,b)=\widetilde{\chi}\bigl(\Delta((a,b))\bigr)$, where $(a,b)=\{w\in P:a<w<b\}$ and $\Delta((a,b))$ is the order complex of this open interval. We want to show that this Euler-characteristic function $F$ satisfies the same recursion as the Möbius function. The recursion we need is \begin{align*} F(u,v)=-\sum_{u\le z<v}F(u,z) \end{align*} for $u<v$. The reason this is the right identity is that the Möbius function on a finite poset is characterized by $\mu_P(u,u)=1$ and \begin{align*} \mu_P(u,v)=-\sum_{u\le z<v}\mu_P(u,z) \end{align*} whenever $u<v$. Start from the right-hand side. Since $F(u,u)=1$, we have \begin{align*} -\sum_{u\le z<v}F(u,z)=-1-\sum_{u<z<v}F(u,z). \end{align*} For each $z$ satisfying $u<z<v$, the definition of reduced Euler characteristic gives \begin{align*} F(u,z)=-1+\sum_{m\ge 1}(-1)^{m-1}c_m(u,z), \end{align*} where $c_m(u,z)$ is the number of chains \begin{align*} u<a_1<\cdots<a_m<z. \end{align*} Substituting this expansion gives \begin{align*} -\sum_{u\le z<v}F(u,z) = -1+\#\{z\in P:u<z<v\} -\sum_{u<z<v}\sum_{m\ge 1}(-1)^{m-1}c_m(u,z). \end{align*} Now we interpret each term as counting chains in the open interval $(u,v)$. The term $\#\{z\in P:u<z<v\}$ is exactly $c_1(u,v)$, because choosing one element $z$ strictly between $u$ and $v$ is the same as choosing a chain \begin{align*} u<z<v. \end{align*} For longer chains, fix $k\ge 2$. A chain \begin{align*} u<z_1<\cdots<z_k<v \end{align*} has a unique last internal element, namely $z_k$. Removing that last element from the endpoint role leaves the prefix \begin{align*} u<z_1<\cdots<z_{k-1}<z_k, \end{align*} which is counted by $c_{k-1}(u,z_k)$. Conversely, any chain counted by $c_{k-1}(u,z)$ for some $z$ with $u<z<v$ extends uniquely to the chain \begin{align*} u<z_1<\cdots<z_{k-1}<z<v. \end{align*} This is a bijection between chains with $k$ internal elements in $(u,v)$ and prefixes counted in the double sum with $m=k-1$. The sign also matches. A prefix with $m=k-1$ contributes $(-1)^{m-1}=(-1)^{k-2}$ inside $F(u,z)$, and the outer negative sign changes this to $(-1)^{k-1}$. Therefore \begin{align*}-\sum_{u\le z<v}F(u,z)=-1+\sum_{k\ge 1}(-1)^{k-1}c_k(u,v).\end{align*} The expression on the right is exactly the reduced Euler characteristic expansion of $F(u,v)$. Hence \begin{align*} F(u,v)=-\sum_{u\le z<v}F(u,z). \end{align*}[/guided]

[step:Identify the auxiliary function with the Möbius function] By the recursive characterization of the Möbius function [citetheorem:8093], the Möbius function $\mu_P$ of the finite poset $P$ is characterized on comparable pairs by $\mu_P(u,u)=1$ and, for $u<v$, $\mu_P(u,v)=-\sum_{u\le z<v}\mu_P(u,z)$. The function $F$ has the same diagonal value and satisfies the same recursion by the preceding step. It remains only to justify uniqueness. Fix comparable $u\le v$. We prove by induction on the cardinality of the interval $[u,v]=\{z\in P:u\le z\le v\}$ that any two functions satisfying the diagonal condition and the recursion agree at $(u,v)$. If $u=v$, both values are $1$. If $u<v$, the recursive formula expresses the value at $(u,v)$ entirely in terms of values at pairs $(u,z)$ with $u\le z<v$, and each interval $[u,z]$ is strictly smaller than $[u,v]$. The induction hypothesis therefore forces the two recursive values to be equal. Applying this uniqueness to $F$ and $\mu_P$ gives $F(x,y)=\mu_P(x,y)$. Since $x<y$, the definition of $F$ gives $F(x,y)=\widetilde{\chi}(\Delta((x,y)))$. Therefore $\mu_P(x,y)=\widetilde{\chi}(\Delta((x,y)))$, which is the desired identity. [/step]