I was reading this post about the Künneth theorem, to try to understand what it is.
I don't really understand what it means for a sequence to be natural. Does this mean we have a natural transformation between some functors? Which ones?
Naturality of an exact sequence
algebraic-topologycommutative-algebraexact-sequencefunctorsnatural-transformations
Related Solutions
Let's setup the framework. Let $\mathsf{Ab}$ be the category of abelian groups, and let $\mathsf{SES}$ be the category of short exact sequences, that is diagrams of the form $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ (with the usual conditions about images and kernels) and morphisms the obvious thing. Then define three functors, $$\begin{align} (H_n^a, H_n^b, H_n^c) & : \mathsf{SES} \to \mathsf{Ab}^3 \\ (A,B,C,f,g) & \mapsto (H_n(A), H_n(B), H_n(C)) \end{align}$$
The functoriality simply means that if you have three exact sequences and two chain maps $X \xrightarrow{\phi} X' \xrightarrow{\psi} X''$, then $H_n^x(\psi \circ \phi) = H_n^x(\psi) \circ H_n^x(\phi)$ for $x = a,b,c,$.
On the other hand the long exact sequence construction is a combination of three families of natural transformations, namely $f_* : H_n^a \to H_n^b$, $g_* : H_n^b \to H_n^c$ (both defined in the obvious way), and the connecting morphism $\partial_* : H_n^c \to H_{n-1}^a$. When one says that the long exact sequence is natural, one is really saying that these three families are all natural transformations.
Of course, you can set things up differently and construct a functor $\operatorname{LES} : \mathsf{SES} \to \mathsf{Ch}$ from the category of short exact sequences to the category of chain complexes, bundling everything together. But when people say that the long exact sequence construction is natural, they are implicitly thinking about all the $H_n$ separately.
This is in fact somewhat of a general construction. If you have two functors $F,G : \mathsf{C} \to \mathsf{D}$ and a natural transformation $\phi : F \to G$, then you can build up a new functor $\mathsf{C} \to \mathsf{Ar}(\mathsf{D})$ given by: $$c \in \mathsf{C} \mapsto (F(c) \xrightarrow{\phi_c} G(c)) \in \mathsf{Ar}(\mathsf{D}),$$ and vice-versa (a functor $\mathsf{C} \to \mathsf{Ar}(\mathsf{D})$ is the same thing as two functors and a natural transformation).
It's hard to say which one is more "natural", but in general using this principle you can interpret any naturality result as a functoriality result, if you want – at the cost of using a more complicated target category.
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.
Best Answer
If $F, G, H:C\rightarrow Ab$ are a functors to an Abelian category. Suppose that for every object $A$ there exists an exact sequence $F(A)\rightarrow G(A)\rightarrow H(A)$. This sequence is natural if and only if for every morphism $f:A\rightarrow B$,
$\matrix{ F(A)&\rightarrow &G(A)&\rightarrow& H(A)\cr F(f)\downarrow &&G(f)\downarrow &&\downarrow H(f) \cr F(B)&\rightarrow &G(B)&\rightarrow& H(B)}$
is commutative.