Claim: Let $A$ be a (not necessarily f.g.) abelian group. Suppose that $T$ is the subgroup of all torsion elements in $A$, and we have that:
$${A}\diagup{T} \cong F$$
where $F$ is a free abelian group. Then:
$$A \cong A \diagup T \oplus F$$
Proof (edited after answer from elemelons)
By the definition of quotient groups, there exists a (set-theoretic) section $s : F \to A$ such that $A = \{s(f) + t : f \in F, t \in T\}$ as a set, where $s$ maps an element $f \in F$ to $a \in A$ such that its coset is $a+T \in A \diagup T$ under the assumed isomorphism.
We claim that there exists a section $s_*$ which is also an injective homomorphism. Let $B = \{f_i\}_I$ be a (not necessarily finite) basis for $F$. Then, we define our section as follows:
$$s_* : F \to A, f \mapsto \sum_{i=1}^N n_i s(f_i)$$
where $f = \sum_{i=1}^N n_i f_i$ for $f_i \in B$, and $s$ is the arbitrary section previously asserted to exist. This is clearly a homomorphism. Suppose that:
$$f = \sum_{i=1}^N n_i f_i, \;\; s_*(f) = \sum_{i=1}^N n_i s(f_i) = 0$$
Then:
$$\sum_{i=1}^N n_i (a_i + T) = 0 + T$$
But from the isomorphism , we know that $A \diagup T$ is free, with $(a_i + T)_I$ forming a basis. Hence, we must have $n_i \equiv 0$.
$$\implies f = 0$$
Thus, $s_*$ is injective. We now consider:
$$A = \{ f + t : f \in F^*, t \in T \}$$
where $F^* := s_*(F) \le A$. Note that $F^* \cong F$, so it is free.
It suffices to show that this sum is direct. Let $f_1, f_2 \in F^*$, $t_1,t_2 \in T$, and suppose that:
$$f_1 + t_1 = f_2 + t_2$$
$$\implies f_1 – f_2 = t_2 – t_1$$
Then, the LHS is an element of a free group, and the RHS is an element of a torsion group. Thus, both must be equal to $0$. Result follows.
Question: Is this proof correct? This all seems quite simple, but I would like to make sure there are no inaccuracies. This result superficially seems to contradict other results (for instance, it is known that not every non-f.g. abelian group can be decomposed in this way), but note that a lot is assumed in the premise of the question.
There are lots of questions on similar topics, but I haven't found a duplicate anywhere.
Best Answer
You refer to elements of $F$ as if they are elements of $A$. But this is not so: there is an isomorphism $A/T\cong F$, so we may identify elements of $F$ with cosets of $T$. We may pick a coset representative $a(f)$ for each such coset, which is essentially a function $a:F\to A$. (We call $a$ a section of the projection $A\to A/T$.) In general, this will not be a homomorphism, so will only identify $F$ with a subset of $A$, not a subgroup. Indeed, without our hypotheses (i.e. $F$ is free), there is no set-theoretic section which is a homomorphism.
To use the hypothesis that $F$ is free, pick a basis for it, pick coset representatives for the basis elements, then "extend linearly" for a section $a:F\to A$ which is indeed additive (and one-to-one; check this). This identifies $F$ with a free subgroup of $A$, and then your proof basically applies at that point.