Q:

Help in understanding proof of Möbius Inversion Formula

Help in understanding proof of Möbius Inversion Formula

Could someone clarify why the equality $$\sum_{d^{'} | (n/d)} \mu(d) f(d^{'}) \sum_{d | n} \mu(d) F(n/d) =  \sum_{d | n} \sum_{d^{'} | (n/d)} \mu(d) f(d^{'})$$, in the below proof, is true ?
Also is there some rigorous way of proving that "As $d$ runs over the positive divisors of $n$ so does $d^{'}$ ?

A:

Let $D_n$ be the set of positive divisors of $n$, for $n \in \mathbb{N}$.
Then you have to show that the function$f: D_n \to D_n$ given by $\displaystyle f(d) = \frac{n}{d}$ is well-defined and bijective.

It's well defined because if $d \mid n$ then $\exists q \in \mathbb{N}: dq=n$, thus $\displaystyle q=\frac{n}{d}$, but $q.d = n$, therefore $q \mid n$, in other words, $\displaystyle \frac{n}{d} \mid n$, hence, $f(n) \in D_n$ and since division is a well-defined arithmetic operation, we see that $f$ is well-defined.
$f$ is injective because if $f(d_1)=f(d_1)$ then $\displaystyle \frac{n}{d_1}=\frac{n}{d_2}$ which gives $d_1 = d_2$.
$f$ is surjective because for any $d \in D_n$, there exists $\displaystyle \frac{n}{d} \in D_n$ such that $\displaystyle f(\frac{n}{d})=d$.

What I just proved is that $f$ relables elements of $D_n$. So, when $d$ runs through the divisors of $n$, $\displaystyle \frac{n}{d}$ does the same. 
But notice that $\displaystyle d' \mid \frac{n}{d}$ if and only if $dd' \mid n$, therefore $dd' \in D_n$.
Now check that the function $g_{d}: D_n \to D_n$ given by $g_{d}(d') = dd'$ is bijective for any $d \in D_n$.
Since the composition of bijections is a third, the function $fog_{d}: D_n \to D_n$ is a bijection, this proves the result you want I guess. I hope that it's clear now.

