Uniqueness of the long exact sequence in homology A few days ago colleagues of mine and I listened to a talk about spectral sequences and one "application" of them was the proof that any short exact sequence (s.e.s.)
$$0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$$
of chain complexes induces a long exact sequence (l.e.s.)
$$\cdots \to H_n(A) \to H_n(B) \to H_n(C) \to H_{n-1}(A) \to \cdots$$
Of course this has little or nothing to do with spectral sequences and is really just the usual diagram chasing proof in disguise and in particular the three kinds of maps that are constructed here are exactly the usual maps $H_n(f)$, $H_n(g)$ and the connecting homomorphism $\delta_n$. The question that then came up was if and how one could prove that these are the maps without doing the diagram chase all over again.
In other words: Is the long exact sequence unique in an appropriate sense? For example is it the only functor (up to natural isomorphism of course) {s.e.s. of chain complexes} $\to$ {l.e.s. of abelian groups} that has $\ldots,H_n(A),H_n(B),H_n(C),H_{n-1}(A),\ldots$ as objects in the long exact sequence?
 A: Suppose that $F$ takes s.e.s. of chain complexes of vector spaces to l.e.s. of vector spaces. Suppose that the objects of $F(0 \to A^{\bullet} \to B^{\bullet} \to C^{\bullet} \to 0)$ are $H^0(A)$, $H^0(B)$, $H^0(C)$, $H^1(A)$, etcetera and that the maps $H^i(A) \to H^i(B)$ and $H^i(B) \to H^i(C)$ are the standard ones. Suppose that $F$ commutes with direct sum and isomorphism. Then $F$ is the standard choice up to choosing a collection of nonzero scalars $c_i$ to rescale the boundary map $H^i(C) \to H^{i+1}(A)$ by. If require further that $F$ commutes with shifts, then we only get one global choice of scalar.
This is because any s.e.s. of complexes of vector spaces is (non-canonically) a direct sum of complexes of  five kinds. Here I write the complexes $A$, $B$ and $C$ horizontally and the maps between them vertically.
$$\begin{pmatrix}
k & \rightarrow & k \\ 
\downarrow & & \downarrow \\
k & \rightarrow & k \\
& & \\
0 & & 0 \\
\end{pmatrix} \quad 
\begin{pmatrix}
0 & & 0 \\
& & \\
k & \rightarrow & k \\ 
\downarrow & & \downarrow \\
k & \rightarrow & k \\
\end{pmatrix} \quad
\begin{pmatrix}
k \\
\downarrow \\
k \\
\\
0 \\
\end{pmatrix} \quad 
\begin{pmatrix}
0 \\
 \\
k \\
\downarrow \\
k \\
\end{pmatrix}\quad
\begin{pmatrix}
0 & & k \\
& & \downarrow \\
k & \rightarrow & k \\ 
\downarrow & &  \\
k &  & 0 \\
\end{pmatrix}$$
So $F$ is determined by how it behaves on these five complexes. On the first two, $H^{\ast}=0$, so $F$ gives the zero complex. On the third and fourth, the only nonzero maps are $H^i(A) \to H^i(B)$ and $H^i(B) \to H^i(C)$ respectively, and we are assuming we know those. So all that we have to do is determine what $F$ does in the fifth case, and this is just a choice of scalar. 
There is a natural choice of scalar, given by composing the horizontal map and the inverses of the vertical maps, but I don't think you can know that your favorite definition uses this natural choice without actually checking it.
