If $n={P_1}^{r_1}.{P_2}^{r_2}.\dots.{P_k}^{r_k}$ then $f(n)={r_1}^{P_1}.{r_2}^{P_2}.\dots.{r_k}^{P_k}$then which one is true?

Question

If $n={P_1}^{r_1}.{P_2}^{r_2}. \dots .{P_k}^{r_k}$ then $f(n)={r_1}^{P_1}.{r_2}^{P_2}.\dots.{r_k}^{P_k}$then which one is true?

1.The function is one-to-one.

2.The function is surjective.

3.If there are $a,b$ that $f(a)=m$,$f(b)=n$ then there is a $c$ that $f(c)=mn$.

4.$f(m).f(n) \ge f(mn)$

5.$f(m).f(n) \le f(mn)$

My attempt:The first one is wrong because both $5^2*11^2$ and $3^2*13^2$ give $2^{16}$.The second one is also wrong because we cannot make primes.The last one is also wrong because consider $m=5^3*3^2$ and $n=5^2*2^3$ which will give $f(5^3*3^2).f(5^2*2^3) \le f(5^5*3^2*2^3)$ which gives $3^5*2^3*2^5*3^2\le 2^3*5^5*3^2$ which gives $6^5 \le 5^5$ which is wrong.But I am stuck in choosing between $3$ and $4$.

Answer 1

The proof of $3$ is easy.

Suppose that $a=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$ and $b=p_1^{b_1}p_2^{b_2}\dots p_k^{b_k}$.

Some exponents may be $0$ so that all of the exponents fit, but we ask that at at least one of $a_i$ and $b_i$ is not zero for each $i$.

Then we have that $f(a)f(b)=g(a_1,b_1)^{p_1}(g(a_2,b_2)^{p_2}\dots g(a_k,b_k)^{p_k}$. Here $g(a_i,b_i)$ is equal to $\max(a_i,b_i,a_ib_i)$

So we have that the number $c= p_1^{g(a_1,b_1)} p_2^{g(a_2,b_2)}\dots p_k^{g(a_k,b_k)}$ satisfies $f(c)=f(a)f(b)$.

Answer 2

(4) is wrong. Note that primes $p$ are mapped to $1$, now let $p$ a prime and $m = n = p$. Then $$ f(m)f(n) = f(p)f(p) = 1 \not\ge f(p^2) = 2^p $$