Composition of multiplicative functions which is not multiplicative

Question

A function $f\colon\mathbb N\to\mathbb C$ is called multiplicative if $f(1)=1$ and $$\gcd(a,b)=1 \implies f(ab)=f(a)f(b).$$ It is called completely multiplicative if the equality $f(ab)=f(a)f(b)$ holds for any pair of positive integers $a$, $b$. (In the definition of multiplicative function we have this condition only for $a$, $b$ coprime.)

If is not difficult to show that if $f\colon\mathbb N\to\mathbb N$ is multiplicative can $g\colon\mathbb N\to\mathbb C$ is completely multiplicative, then $g\circ f$ is multiplicative. (We just notice that if $\gcd(a,b)=1$ then $g(f(ab))\overset{(*)}=g(f(a)f(b))\overset{(\triangle)}=g(f(a))g(f(b))$, where the equivality $(*)$ follows from the fact that $f$ is multiplicative and in the equality $(\triangle)$ we use that $g$ is completely multiplicative.)

What are some simple examples showing that the word completely cannot be omitted, i.e., examples of multiplicative functions $f$ and $g$ such that the composition $g\circ f$ is not multiplicative?

Answer 1

I can think of one. Take $f=g$ with, for any $a \in \mathbb{N},$ $f(a) = |\{k \in \mathbb{N}, k | a\}|$ (number of divisors including himself)

$f\circ g (2 \times 3) = f (4) = 3$

$f\circ g (2) = 2 $

$f\circ g (3) = 2 $

Answer 2

An elementary example: take two distinct primes $p,q$. Then $\mu(\tau(pq))=0$ while $\mu(\tau(p))\mu(\tau(q))=1$, where $\tau(n)$ denotes the number of positive divisors of $n$.

Answer 3

If one would choose the identity for $f$, and $g$ would be a function returning the product of all single prime factors of the argument (removing duplicates), those should both be multiplicative (at least it seems to me), but for example $2=g(8)=g(f(2\cdot4))\neq g(f(2))g(f(4))=2\cdot2=4$