How to understand the naturality in the homology long exact sequence? I have this confusion when reading some notes.
Theorem. Let  $0_\star\rightarrow A_\star\overset{f}{\rightarrow }B_\star\overset{g}{\rightarrow }C_\star\rightarrow 0_\star$ be a short exact sequence of chain complexes.  Then there is a natural homomorphism $\partial:H_n(C)\rightarrow H_{n-1}(A)$ s.t. the following sequence is exact.



I would like to understand the naturality here. It is probably in the sense of natural transformation. But I would like to see what are the functors from what to what, in order to regard this $\partial$ as a natural transformation.
 A: Let $\mathcal {Chain}$ the the category of chain complexes of abelian groups. $\mathcal{Chain}$ is an abelian category.
Let $\mathcal{ShExact}$ be the category of short exact sequences of objects in $\mathcal{Chain}.$
There are forgetful functors $A,B,C:\mathcal{ShExact}\to \mathcal{Chain},$ sending $0\to a_*\to b_*\to c_*\to 0$ to $a_*, b_*,c_*,$ respectively.
There are homology functors $H_i:\mathcal {Chain}\to \mathcal{Ab}.$
Then this is a natural function $\partial:H_{i}\circ C\to H_{i-1}\circ A$ for each $i.$
Of course, the rest of the long exact sequence is also natural. $H_i\circ A\to H_i\circ B$ and $H_i\circ B\to H_i\circ C.$ Those are more obviously existing and natural.

We can avoid the indexes by thinking of homology as a single functor $H$ going from $\mathcal {Chain}$ to $\mathcal C=\mathcal {Ab}^{\mathbb N}.$  Then you need to add a shift functor on $\mathcal C,$ with $S(A_0,A_1,\dots)=(0,A_0,A_1,\dots).$
Then you have natural functions:
$$\alpha: H\circ A\to H\circ B\\
\beta: H\circ B\to H\circ C\\
\partial: H\circ C\to S\circ H\circ A.$$
Again, only $\partial$ is surprising.
A: Naturality says that if you have a map short exact sequences of complexes
$S\to S'$ i.e. a diagram of the form
$\require{AMScd}$
\begin{CD} 
0 @>>> X @>i>> Y @>\pi>> Z @>>> 0\\ 
{} @VfVV   @VgVV @VhVV \\
0 @>>> X'  @>>i'>Y' @>>\pi'> Z' @>>> 0
\end{CD}
the following diagram commutes for each $n\in\mathbb Z$:
\begin{CD} 
 H_n(Z) @>>\delta_S> H_{n-1}(X) @>>>H_n(Y) @>>> H_n(Z) \\ 
  @VVV   @VVV @VVV@VVV\\
H_n(Z') @>>\delta_S> H_{n-1}(X') @>>>H_n(Y') @>>> H_n(Z') \\ \end{CD}
You can think of this as an infinite "staircase" all whose squares commute. I think the other answers have more-or-less stated things in purely categorical terms: the LES can be thought of a functor that takes a short exact sequence of complexes (in some abelian category) and sends it to a long exact sequence of objects (in the same abelian category), with maps $(Hi,Hp,\delta)$. A morphism of short exact sequences $(f,g,h)$ is then sent to a morphism of long exact sequences $(Hf,Hg,Hh)$.
A: A possible interpretation is that this is a natural transformation between functors with domain in the category of short exact sequences of complexes.
