# 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.

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”.

• Thank you for your answer, however, I don't think this really answers the question. I understood what the author was claiming, I just did not understand why it was true. In particular, the second instance of "(prove it)" is exactly the question. I think I might be able to prove it if I had answers to Confusion 1 and 2 in my question. In particular, how are the morphisms mapped by $F$ (I understand how the objects are mapped by $F$), and how can naturality of the map $A \to A/TA$ be implied from $A \cong TA \oplus (A/TA)$ when we are not given a categorical definition of $\oplus$. Nov 7, 2020 at 16:00
• I'm sorry, I misread the question. I added the description of the action of $F$ on objects, so now it should be easier for you to prove that $\theta$ is natural (just write down the naturality square). I don't entirely understand what your “Confusion 2” is about: can you explain it better? Nov 7, 2020 at 16:50
• You actually implicitly answered my "Confusion 2" by explaining how the morphisms are mapped I think. Thank you Nov 7, 2020 at 17:29