Why $f$ is injective? (infinitude of primes) In this very short paper by  Dustin J. Mixon, I would like understand why the author says 

$f$ is injective by the fundamental theorem of arithmetic.

In my opinion, the Fundamental Theorem of Arithmetic (FTA) is necessary to define $f$, but it isn't necessary to prove that $f$ is injective. For example, if FTA were not true but you were able to ensure the uniqueness of $k_i$ by any other way then $f$ would be injective because $(k_1,\ldots,k_N)=(m_1,\dots,m_N)\Rightarrow k_i=m_i$.
In other words, I think the uniqueness of $k_i$ (and therfore FTA) is necessary to define $f$, but not to prove that $f$ is injective.
What can you talk about this?
Thanks.
 A: The fundamental theorem of arithmetic isn't necessary to define $f$, nor is it necessary to prove the injectivity of $f$.
The injectivity is immediate because
$$ g \colon \lbrace 0,\, \ldots,\, K\rbrace^N \to \mathbb{N}; \; g(e_1,\, \ldots,\, e_N) = \prod\limits_{i = 1}^N {p_i}^{e_i}$$
is a left inverse of $f$.
To define $f$, all you need is that each number has some representation as a product of primes. If the FTA didn't hold, such a representation would in general not be unique, but for a finite set of numbers and primes, a representation could be arbitrarily chosen even without thinking about the axiom of choice.
A: Existence of a prime factorization of each positive integer is necessary for existence of $f(x)$, and uniqueness of a prime factorization of each positive integer is necessary for $f(x)$ to be well defined.
Injectivity of $f$ does not depend on either existence or uniqueness of prime factorizations.  It just says there can't be more than one value of $x$ that is equal to a specified product of other numbers.
