I'm trying to understand this paper: Rota, Gian-Carlo. "On the foundations of combinatorial theory I. Theory of Möbius functions." Or at least, to understand the first few pages.
In particular, I'm having some trouble understanding the claim that the möbius inversion formula does in fact provide the inverse of the zeta function. [That is, inverse in the incidence algebra where composition is by convolution and the identity is the Kronecker delta].
So we have a locally finite partially ordered set, and we define $\zeta(x,y) = 1$ if and only if $x\le y$. Also define $\delta(x,y) = 1$ iff $x=y$. We want to find the $\mu$ such that
$$ \sum_{x\le z\le y} \zeta(x,z)\mu(z,y) = \delta(x,y) $$
Rota claims that the $\mu$ that does the trick is defined inductively as follows: $\mu(x,x) =1$ for all $x$, and if $\mu$ has been defined for all $z$ such that $x\le z <y$, then
$$ \mu(x,y) = -\sum_{x\le z<y} \mu(x,z) $$
This is well defined since the poset is locally finite.
Rota says "Clearly $\mu$ is an inverse of $\zeta$" but I'm struggling to see how this is at all clear. Could someone give me a hint as to how the proof of this claim would go?
Suppose you have a function $f$ in your incidence algebra, you want the inverse, i.e. a function $f^{-1}$ such that $f(x,x)f^{-1}(x,x)=d(x,x)=1$, it means $f^{-1}(x,x)=\frac{1}{f(x,x)}$. And $$\sum _{x\leq z\leq y}f(x,z)f^{-1}(z,y)=\sum _{x<z\leq y}f(x,z)f^{-1}(z,y)+f(x,x)f^{-1}(x,y)=d(x,y)=0,$$ so, by rearranging the terms, $$f^{-1}(x,y)=\frac{1}{f(x,x)}(-\sum _{x<z\leq y}f(x,z)f^{-1}(z,y)).$$ Use $\zeta$ instead of $f$.
Edit:
As the OP and Nishant noticed, the restriction in the sum on the Möbius recursion is $x\leq z < y$, for that is to do the same thing done below but using $f^{-1}$ on the left of the convolution, i.e. $$\sum _{x\leq z \leq y}f^{-1}(x,z)f(z,y)=\sum _{x\leq z < y}f^{-1}(x,z)f(z,y)+f^{-1}(x,y)f(y,y).$$It can be done by the fact that Dirichlet convolution is associative.