Direct proof that infinite product of copies of $\mathbb{Z}$ is not projective

Question

It is well-known that the abelian group $$A = \prod_{n=1}^\infty \mathbb{Z}$$ is not free (see, for example this MO question), and that over a PID being free is equivalent to being projective (see here), but the latter uses the axiom of choice. Is there a direct way to see that $A$ is not projective in terms of the lifting property, ideally not using choice?

Answer 1

Here is a proof that does not use the axiom of choice. Let $e_n\in A$ denote the sequence whose $n$th term is $1$ and all other terms are $0$; we take it as known that any homomorphism $A\to\mathbb{Z}$ which vanishes on each $e_n$ is $0$ everywhere (the standard proofs of this certainly do not use choice).

Suppose $A$ were projective. Let $F$ be the free group on the underlying set of $A$, and let $p:F\to A$ be the canonical epimorphism. Since $A$ is projective, there is a splitting $i:A\to F$. Now note that for each $a\in A$, there is a homomorphism $\pi_a:F\to\mathbb{Z}$ which takes a formal linear combination of elements of $A$ and gives you the coordinate of $a$. For any $x\in F$, there are only finitely many $a\in A$ such that $\pi_a(x)\neq 0$. In particular, there are only countably many $a\in A$ such that $\pi_a(i(e_n))\neq 0$ for some $n$. But since $\pi_a\circ i:A\to\mathbb{Z}$ vanishes iff it vanishes on each $e_n$, this means that $\pi_a\circ i=0$ for all but countably many $a$. This means that actually the image of $i$ is contained in the free group $G\subset F$ on a countable subset of $A$. But this $G$ is then countable, which is a contradiction since $A$ is uncountable.

(Here I implicitly used the fact that a countable union of finite sets is countable, which requires choice in general. However, the finite sets are finite subsets of $A$, and since it is possible to totally order $A$, we can use this to canonically biject each of the finite subsets with a finite ordinal and thus conclude that their union is countable.)