Proofs for Multiplicative Functions in Number Theory Part 1:
A theorem in my book proves that if $f(n)$ is a multiplicative function, and $g(n)=\sum_{d|n}f(d)$, then $g(n)$ is also multiplicative. How do I prove the converse of this. The converse being that if $g(n)$ is a multiplicative function, and $g(n)=\sum_{d|n}f(d)$, then $f(n)$ is multiplicative.
Part 2:
This second part goes hand-in-hand with part 1. How do I prove that if $g(n)$ is a multiplicative function, then there is exactly one function $f(n)$ for which $g(n)=\sum_{d|n}f(d)$, and $f(n)$ is a multiplicative function.
What I thought:
For part 1, it looks to me that I might have to use Möbius inversion. I'm not sure though. For part 2, it looks to me that I may have to use induction to prove it. 

For part I when you use Mobius inversion you get :
$$f(n)=\sum_{d|n}\mu(n/d)g(d) $$
Now because $\mu$ and $g$ are both multiplicative, $f$ will be multiplicative as well. This is a slightly more general version of your property that :
$$n\mapsto\sum_{d|n}h(n/d)g(d) $$
is a multiplicative function when $h$ and $g$ both are.
For part II there is no need for induction once you use Moebius inversion formula. If $f_1$ and $f_2$ are two functions such that :
$$\sum_{d|n}f_2(d)=g(n)=\sum_{d|n}f_1(d) $$
Then you get that :
$$ f_2(n)=\sum_{d|n}\mu(n/d)g(d)=f_1(n)$$
