Exactness and Naturality

Question

I'm trying to read this blog post about exact functors, and I see mentions of naturality which I have not stumbled upon elsewhere. In particular, in the proof of the Theorem, the author says

By exactness we have a natural isomorphism...

I searched and found this related post in which one of the answers said the naturality becomes obvious given enough mathematical maturity. I do not have enough, so I need this to be spelled out for me (naturality is important particularly because of this article cited in the blog post).

How, generally speaking, are natural transformations and exact sequences related in the same sense as in the blog post?

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Added by request: Definitions I'm working with: The kernel of an arrow is its equalizer with the zero arrow and its cokernel is its coequalizer with the zero arrow. The image of an arrow is any monic from the epi-mono factorization $f=me$. (Sometimes one talks the image as an equivalence class monics) A composable pair of arrow $\bullet \overset{f}{\rightarrow}b\overset{g}{\rightarrow} \bullet$ is said to be exact at $b$ if $\mathrm{im}f \equiv \mathrm{ker}g$, where $\equiv$ denotes equivalence of subobjects (having isomorphic domains).

Answer 1

Well, as other sources and the comments suggests, by proper usage of the universal properties, one can avoid speaking about natural transformations such as $\def\im{{\rm im\,}}\def\tto{\dashrightarrow} \im f\tto \ker g$ for exact sequences $\overset f\to \overset g\to$.

However, probably the following lemma and preliminary setting can be useful to exactly follow the thought.

Let me emphasize that the exactness is used only for stating the isomorphisms. For the naturality part we basically only talk about functors.

Recall that the objects of the arrow category $\mathcal A^\to$ of $\mathcal A$ are the arrows of $\mathcal A$ and the morphisms between them are the commutative squares.
Lemma: Let $F,G:\mathcal A\to\mathcal B$ be functors. Then, a mapping $\varphi_0:Ob\mathcal A\to Mor\mathcal B$ is a natural transformation $F\tto G$ iff it extends to a functor $\varphi:\mathcal A\to\mathcal B^\to$ such that ${\rm dom}\circ\varphi=F$ and ${\rm cod}\circ\varphi=G$.

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First, for each arrow $f:A\to B$ in our Abelian category $\mathcal A$, $\,$fix a kernel $\ker f\hookrightarrow A$ and an image $ \im f\hookrightarrow B$. This way we obtain functors on the arrow category $\mathcal A^\to\to\mathcal A^\to$, which can also be seen as natural transformations $\ker\dashrightarrow {\rm dom}$ and $\im\dashrightarrow {\rm cod}$ between functors $\mathcal A^\to \to \mathcal A$.

Similarly, we fix kernels and images in $\mathcal B$ before anything happens.

Now, consider also the category $\mathcal A^{\to\to}$ of consecutive arrows in $\mathcal A$. Then the following mappings extend to functors (are natural):

• $U:=\underset{X\to Y}\alpha\ \mapsto \ (k,\alpha)\ \mapsto \ (F(k),F(\alpha))\ \mapsto F(k)\ \mapsto\ \im F(k)$
  
  $\phantom{U:=} \ \mathcal A^\to \to \mathcal A^{\to\to}\ \ \to \quad\quad \mathcal B^{\to\to}\quad\quad\to \ \ \mathcal B^\to\quad\to\ \ \mathcal B$
  
• Similarly, $V:=\alpha\ \mapsto \ker F(\alpha)$.

Finally, there are unique isomorphisms $\ker F(\alpha)\cong \im F(k)$ and $\im F(k)\cong F(\ker\alpha)$ making the obvious triangles commutative, which define the natural isomorphisms such as $U\tto V$.