User:Metajellyfish/Math720/HW8

\item Suppose $\Pi$ is a poset and $f, g : \Pi \to \mathbb{C}$. Then $$f(x) = \sum_{y \leq x} g(y) \iff g(x) = \sum_{y \leq x} \mu (x,y) f(y)$$ Proof:

The expression, $$f(x) = \sum_{y \leq x} g(y) = \sum_{x\in \Pi} \zeta (x,y) g(x)$$ implies that $f = \zeta (g)$ which, coupled with $\mu = \zeta^{-1}$, gives us $\mu(f) = g$:\\ $$\implies \mu (f(x)) = \mu ( \sum_{x \in \Pi} \zeta (x,y) g(y))$$ $$\implies g(x) = \sum_{x\in \Pi} \mu(x,y) f(y) = \sum_{y \leq x} \mu (x,y) f(y)$$ by Lemma 2.2(b). Therefore, $$g(x) = \sum_{y \leq x} \mu (x,y) f(y)$$