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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?

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    $\begingroup$ The connecting morphism is not quite unique. You could replace $\delta_n$ with $-\delta_n$, for example. So there is a genuine ambiguity. Regardless, the long exact sequence (including the connecting morphism) can be defined in a very natural way one you have the notion of mapping cone. $\endgroup$ – Zhen Lin Nov 27 '14 at 20:00
  • $\begingroup$ You might be interested in the notion of a "left derived functor". $\endgroup$ – user14972 Nov 27 '14 at 20:06
  • $\begingroup$ @ZhenLin: I know that. The long exact sequence where $\delta$ is replaced by $-\delta$ is naturally isomorphic to the usual one. $\endgroup$ – Johannes Hahn Nov 27 '14 at 20:34
  • $\begingroup$ @Hurkyl: I know the concept. How exactly does it help in this case? $\endgroup$ – Johannes Hahn Nov 27 '14 at 20:35
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    $\begingroup$ I doubt you could get away with that. Actually, I doubt you could avoid looking at the construction of spectral sequences. (The "proof" with spectral sequences is almost surely circular. You need the long exact sequence in homology in order to construct the spectral sequence of a filtered complex.) $\endgroup$ – Zhen Lin Dec 9 '14 at 9:27
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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.

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  • $\begingroup$ Do you know if something similar could be attempted over a general ring (maybe just a PID)? $\endgroup$ – Najib Idrissi Dec 16 '14 at 13:00
  • $\begingroup$ Over a general ring, I doubt it. Classifying free complexes of free modules over $R$ should be strictly harder than classifying modules, and that's way too hard. For a PID, I would guess yes but I don't know it for sure. $\endgroup$ – DES-SupportsMonicaAndTransfolk Dec 16 '14 at 16:44

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