Problem
Let $f : \mathbb{N}\to\mathbb{N}$ be a strictly increasing function such that $f(2) = 2$ and $f(mn) = f(m) \cdot f(n)$ for every relatively prime pair of positive integers $m$ and $n$. Prove that $f(n) = n$ for every positive integer $n$.
My approach
I tried to prove this using induction. For the base case, we need to find $f(3)$. We have $$f(3) \cdot f(5)<2f(9)<4f(5)$$ $$\implies f(3)<4$$ And since $f(3)>f(2)=2$, we have $f(3)=3$.
Now, we assume that $f(k-1)=k-1$ and $f(k)=k$. Since $k$ and $k-1$ are relatively prime, we have $f(k(k-1))=f(k)f(k-1)=k(k-1)$. But this does not prove.
So, how do I complete the proof?
Show this little lemma previously: If $f:\mathbb{N}\rightarrow \mathbb{N}$ is a strictly increasing function over naturals and we have $a,b\in \mathbb{N}$ such that $f(a) = a, f(b) = b$ then $f(x) = x $ for all naturals in the interval $[a,b]$.
Proof of the lemma: Let $S$ be the set of naturals $S = \{a+1, ..., b-1\}$. Because $f$ is strictly increasing, we have $a < f(a+1) < f(a+2) < ... < f(b-1) < b$. In particular this means $f(S) = S$. Furthermore, $f(a+1)$ is the minimum of all values, because all those must be different it must mean that $f(a+1) = a+1$ because that's the minimum of $S$. If it wasn't $a+1$ you can reach a contradiction, because there must be a number in $S$ with image $a+1$. You can finish by applying this result to $f(a+1) = a+1, f(b) = b$.
After that you just apply your induction solution with $f(2) = 2$ and $f(k(k-1)) = k(k-1)$.