Why would a natural isomorphism $A \cong TA \oplus (A / TA)$ imply that $A \twoheadrightarrow A/TA \rightarrowtail TA \oplus (A / TA)$ is natural? $\newcommand{\abcat}{\text{Ab}_\text{fg}}$
$\newcommand{\tgroup}{TA \oplus (A/TA)}$
$\newcommand{\epi}{\twoheadrightarrow}$
$\newcommand{\mono}{\rightarrowtail}$
Let $A$ be an object in the category $\abcat$ of finitely generated abelian groups. And let $TA$ denote its torsion subgroup.
In Category Theory in Context the author proves that the isomorphisms $A \cong \tgroup$ are not natural in proposition 1.4.4.
The proof starts out by stating the following (note, $\epi$ denotes an epimorphism, and $\mono$ denotes a monomorphism)

Suppose the isomorphisms $A \cong \tgroup$ were natural in $A$. Then the composite
$$A \epi A/TA \mono \tgroup \cong A$$
of the canonical quotient map, the inclusion into the direct sum, and the hypothesized natural isomorphism would define a natural endomorphism of the identity functor on $\abcat$

My main question is simple: "why?".
But I have a few confusions which may relate to why I am having trouble with the main question.
Confusion 1
A natural transformation is between functors $F, G : C \rightrightarrows D$, so what exactly are the functors in the proposed natural isomorphism? I am guessing that one of the functors $F$ is the identity functor on $\abcat$. Then perhaps the other functor $G$ is an endomorphic functor on $\abcat$  where the action on an object of $\abcat$ is $$A \mapsto \tgroup$$ but then how are the morphisms mapped by the functor? Earlier the author does state,

In practice, it is usually most elegant to define a natural transformation by saying that the arrows $X$ are natural, which means that the collection of arrows defines the components of a natural transformation, leaving implicit the correct choices of domain and codomain functors, and source and target categories.

But as I am new to this, I am not sure how these "correct" choices are "implicit". It may also be the case that the author is proving that there are no functors which have a natural isomorphism between them and also map the objects of $\abcat$ in the way described above. Is that what is going on here?
Confusion 2 If the functor $G$ really does map the objects $A$ to the objects $\tgroup$ then I fail to see how the isomorphism $A \cong \tgroup$ has anything to do with a natural epimorphism $A \epi A/TA$ or a natural monomorphism $A /TA \mono \tgroup$ from a categorical perspective. I have a foggy feeling that it is indeed true intuitively. However, I don't understand how a natural transformation could imply this using only category theory formally. (Unless I missed it, the author has not defined $\oplus$ using category theory yet, would this be necessary?)
I am assuming by the way this first part of the proof was worded and also by the previous proofs and examples in the book that we only need a simple categorical argument here, and that neither a group theoretical argument (appealing to the elements of the group) or an extremely complicated categorical argument is necessary. But I fail to see what the argument is.
 A: Let $\mathcal{A}$ be the category of finitely generated abelian groups. The mapping $A \mapsto TA \oplus (A/TA)$ extends to an endofunctor $F: \mathcal{A} \to \mathcal{A}$ as follows: take a morphism $f: A \to B$ of finitely generated abelian groups. Construct the morphism $Ff: TA \oplus (A/TA) \to TB \oplus (B/TB)$ as follows:

*

*$f$ maps torsion elements to torsion elements (if $n \cdot a = 0$, then $n \cdot f(a) = f(n \cdot a) = 0$, so $n \cdot a$ is torsion), so $f$ induces a morphism $f|_{TA}: TA \to TB$ by restriction;

*There is a morphism $g: A/TA \to B/TB$ defined by putting $g(a + TA) = f(a) + TB$. This well-defined: if $a + TA = a' + TA$, then $a - a'$ is in $TA$ and $f(a - a') = f(a) - f(a')$ is in $TB$ by the previous observation, so
$$g(a + TA) = f(a) + TB = f(a') + TB = g(a' + TA)
$$
and $g$ is well-defined. Notice that $g$ is also a morphism of groups because $f$ is;

*now we put $Ff = f|_{TA} \oplus g$. That is, an element $(a, a' + TA)$ of $TA \oplus (A/TA)$ gets mapped to $(f(a), f(a') + TB)$ in $TB \oplus (B/TB)$ by $Ff$.

Now it is quite easy to prove that $F$ is indeed a functor. What I did might look complicated, but it is indeed quite tautological: “taking torsion” is a functor $T: \mathcal{A} \to \mathcal{A}$, “modding out by torsion” is a functor $(-)/T(-): \mathcal{A} \to \mathcal{A}$ and “taking direct sums” is a functor $\oplus: \mathcal{A} \times \mathcal{A} \to \mathcal{A}$: by appropriately assembling those three functors you get $F$.
Then saying that there are natural isomorphisms $A \cong TA \oplus (A/TA)$ means that there is a natural isomorphism $\eta: F \Rightarrow \mathsf{id}_{\mathcal{A}}$ between $F$ and the identity endofunctor (think about it: this consists of a family of isomorphisms $\eta_A: TA \oplus (A/TA) \cong A$ for each object $A$ of $\mathcal{A}$).
Now there is a natural transformation $\theta: \mathsf{id}_{\mathcal{A}} \Rightarrow F$ such that for a finitely generated abelian group $A$, the morphism $\theta_A: A \to TA \oplus (A/TA)$ is the composition $A \to A/TA \to TA \oplus (A/TA)$ (prove it). If $\eta$ as above exists, you can consider the composition $\eta \circ \theta: \mathsf{id}_{\mathcal{A}} \Rightarrow F \Rightarrow \mathsf{id}_{\mathcal{A}}$, which is explicitly the composition $A \to A/TA \to TA\oplus A/TA \to A$ (where the last morphism is $\eta_A$). Since $\eta$ is supposed natural (by contradiction) and $\theta$ is natural, the composite $\eta \circ \theta$ must be a natural endomorphism of the identity functor, and this is exactly what the author means by “the hypothesized natural isomorphism would define a natural endomorphism of the identity functor”.
