Is there a functor $F$ preserving finite direct sums but not split exact sequences, for which $F\mathbb{Z}$ is free and finitely generated?

Question

Is there a functor $F$ from the category of abelian groups to itself, such that

1. $F$ preserves finite direct sums (as well as the empty direct sum) up to (arbitrary) isomorphism
2. $F$ doesn't in general preserve split exact sequences
3. $F\mathbb{Z}$ is free and finitely generated.

By 2., I mean that there should exist abelian groups $A$ and $B$ such that the sequence $$ 0\to FA\overset{Fi}{\to}F(A\oplus B)\overset{Fp}{\to}FB\to 0 $$ isn't exact, where $i:A\to A\oplus B$ is the canonical injection and $p:A\oplus B\to B$ is the canonical surjection.

In this question, I asked whether there is a functor $F$ from the category of abelian groups to itself satisfying 1. and 2., and after some time I found such a functor, the construction of which I posted also in the original question. The basic idea is to first quotient out the torsion, i.e. sending $A$ to $GA:=A/\operatorname{Tor}(A)$, and then sending $GA$ to the free abelian group $FA:=\mathbb{Z}^{\oplus GA}/\mathbb{Z}\cdot\delta_0$, where $\delta_0$ is the element in $\mathbb{Z}^{\oplus GA}$ which has a coefficient of $1$ at $0=0_A$, but is $0$ everywhere else. One can check that then $F(A\oplus B)$ and $F(A)\oplus F(B)$ are free abelian of equal rank, for all $A,B$.

However, in this construction $F(\mathbb{Z})$ is infinitely generated, and furthermore the construction heavily relies on sending abelian groups to infinitely generated ones, in order to exploit that on infinite cardinals, multiplication and addition are equal. In the context where this question came up, it would be interesting to know whether there is such a functor, but I think it is also interesting of itself.

Answer 1

Fix an uncountable set $X$. For any set $S$, let $G(S)$ denote the set of injective functions $X\to S$. Define $F(A)$ to be the free abelian group on $G(A)$. We can make this $F$ into a functor: a homomorphism $f:A\to B$ induces a partial function $G(A)\to G(B)$ by composition (partial because some injections $X\to A$ may no longer be injective after composition with $f$), and then we can get a homomorphism $F(f):F(A)\to F(B)$ by mapping the generators on which $G(A)\to G(B)$ is undefined to $0$.

This functor has $F(\mathbb{Z})=0$ which is free and finitely generated. The fact that $F$ preserves finite direct sums up to isomorphism follows easily from the fact that $F(A)$ is trivial unless $|A|\geq |X|$, in which case $F(A)$ is free of rank $|A|^{|X|}$ (and these cardinals are infinite). Finally, $F$ does not preserve split exact sequences since you can have in injection $X\to A\oplus B$ which is not injective when projected to $B$ but does not have its image contained in $A$, and this will give an element of the kernel of $F(A\oplus B)\to F(B)$ which is not in the image of $F(A)\to F(A\oplus B)$.