# Construction of the morphism from the zig-zag lemma

UPD: I'm not sure why i'm not getting any comments or votes, so I'm expanding a little bit below to make it easier to understand my question and make it more self-contained.

For reference I'm using the old edition of Lang's "Algebra", the one with homologies in their own chapter, but I'm sure it's there in the 2002 edition, too.

My question is about construction of the morphism that is the subject of the zig-zag lemma. Let $A$ be a ring, and consider three chain complexes of $A$-modules $(E', d)$, $(E, d)$ and $(E'', d)$ forming a short exact sequence $$0 \to E' \ \xrightarrow{f} \ E \ \xrightarrow{g} \ E'' \to 0.$$ From now on, following Lang, I will be dropping indices and parentheses where possible to make notation lighter. I will also denote submodules of cycles with a letter $Z$ and boundaries with the letter $B$ as usual. The homologies will be considered as graded modules given by $H_i(E) = Z_i / B_i$ and so on.

The problem is to prove that the formula $z'' \mapsto f^{-1} d g^{-1} z''$ defines a morphism $$\delta: H(E'') \to H(E'), \qquad \delta_i: Z''_i / B''_i \to Z'_{i+1} / B'_{i+1}.$$

Below is my completion of Lang's half page sketch of the explicit construction together with the proof of its correctness. The actual question is in the end.

Let $z'' \in Z''_i$. Then the set $f^{-1} d g^{-1} z'' \subset E'_{i+1}$ is not empty. Indeed, components of $g$ are surjective, so there is a $z \in E_i$ such that $gz = z''$. Since $0 = dz'' = dgz = gdz$, we have $dz \in \ker g = \operatorname{im} f \subset E_{i+1}$, therefore there is a $z' \in E'_{i+1}$ such that $fz' = dz$, and thus $f^{-1} d g^{-1} z'' \subset E'_{i+1}$ is not empty. Moreover, $f^{-1} d g^{-1} z'' \subset Z'_{i+1}$. Indeed, for all $z \in E_i$ such that $gz = z''$ and $z' \in E'_{i+1}$ such that $fz' = dz$ we have $fdz' = dfz' = d^2z = 0$.

Next, we prove that for all $z'_1, z'_2 \in f^{-1} d g^{-1} z''$ we have $z'_1 - z'_2 \in B'_{i+1}$, or in other words that there is some $t \in E'_i$ such that $dt = z'_1 - z'_2$. We pick $z_1, z_2$ just like before, and observe that $f(z'_1 - z'_2) = d(z_1 - z_2)$. Since $g(z_1 - z_2) = z'' - z'' = 0$, then there is $t \in E'_i$ such that $ft = z_1 - z_2$, and we have $fdt = dft = d(z_1 - z_2) = f(z'_1 - z'_2)$. But every component of $f$ is injective, so we indeed have $dt = z'_1 - z'_2$.

Thus we have a mapping $\gamma_i: Z''_i \to Z'_{i+1} / B'_{i+1}$, $z'' \mapsto f^{-1} d g^{-1} z'' + B'_{i+1}$. We need to prove that it is a homomorphism. Let's prove that $\gamma_i(z''_1 + z''_2) = \gamma_i(z''_1) + \gamma_i(z''_2)$. Indeed, let's pick $z_1, z_2 \in E_i$ such that $gz_1 = z''_1$, $gz_2 = z''_2$. Then $g(z_1 + z_2) = z''_1 + z''_2$, and $d(z_1 + z_2) = d(z_1) + d(z_2)$. Let's pick $z'_1, z'_2 \in Z'_{i+1}$ such that $fz'_1 = dz_1$, $fz'_2 = dz_2$. Then $f(z'_1 + z'_2) = d(z_1 + z_2)$, thus if $z'_1 \in f^{-1} d g^{-1} z''_1$ and $z'_2 \in f^{-1} d g^{-1} z''_2$, then $z'_1 + z'_2 \in f^{-1} d g^{-1} (z''_1 + z''_2)$. But the classes of $z'_1, z'_2, z'_1 + z'_2$ modulo $B'_{i+1}$ is independent of the choice of $z_1, z_2$, and obviously $z'_1 + z'_2 + B'_{i+1} = (z'_1 + B'_{i+1}) + (z'_2 + B'_{i+1})$, and thus indeed $\gamma_i(z''_1 + z''_2) = \gamma_i(z''_1) + \gamma_i(z''_2)$. In the same way we prove that $\gamma_i(\alpha z') = \alpha \gamma_i(z')$ (I haven't actually done this, but it seems like it go the same way). Thus $\gamma_i$ is indeed a homomorphism.

Now all that's left to prove is that $B''_i \subset \ker \gamma_i$. Indeed, consider a $z'' \in B''_i$. Then there is a $w'' \in E''_{i-1}$ such that $dw'' = z''$, and a $w \in E_{i-1}$ such that $gw = w''$. Thus $z'' = dgw = gdw$. Thus there is a $z \in B_i$ such that $z = dw$ and $gz = z''$. Obviously, $dz = 0$, and because every component of $f$ is an injection, there is only one $z'_0 \in Z'_{i+1}$ such that $fz'_0 = dz = 0$, and it is $z'_0 = 0$. But then for all $z' \in f^{-1} d g^{-1} z''$ we have $z' - z'_0 = z' \in B'_{i+1}$, and thus $f^{-1} d g^{-1} z'' \subset B'_{i+1}$, as was to be proved.

Now by the homomorphism theorem $\gamma_i$ uniquely factors through some $\delta_i: Z''_i / B''_i \to Z'_{i+1} / B'_{i+1}$, and this is what we needed to construct.

My two questions are:

1. Is this proof correct and complete?
2. Was there a shortcut that I missed?

UPD: I understand that this is not a delightful question to attempt to answer. So God giveth, God taketh: I'll award the bounty to the most serious flaw found, or to Dylan if none are found.

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This looks good to me! – Dylan Moreland Aug 17 '11 at 19:53
+1 for taking time to actually check this :) – Alexei Averchenko Aug 17 '11 at 20:48
I didn't see this question before and I agree with Dylan. Your proof is correct but, as you will be certainly be aware of, not quite complete, since you didn't check that $\gamma$ commutes with scalar multiplication. I don't think there is any significant short cut (using this approach). You should now check that you get a long exact sequence in homology... – t.b. Aug 19 '11 at 14:09

Since I think an answer should be more informative than this, I've uploaded a proof of the snake lemma for chain complexes of vector spaces at Scribd. These are notes I took on a lecture given by Kevin Costello — any errors are surely mine. His proof is slightly more conceptual and you might like it. Unfortunately, I haven't had time to think about whether this can be adapted to the general case; in particular, at some point he chooses a splitting of each $0 \to A^i \to B^i \to C^i \to 0$ and this is often unavailable in the general case.
The proof in your notes is neat but unfortunately it doesn't really generalize directly. However, what you can do is use the mapping cone of the map $\alpha: A \to B$ and that gives a degree-wise split sequence $B \to C(\alpha) \to A[1]$ to which you can apply the argument in your notes. Furthermore there is a natural cochain map $C(\alpha) \to C$ which is a quasi-isomorphism (isomorphism in homology) which is in the end responsible for he long exact seq. But fleshing out all this is certainly as long as Alexey's argument. – t.b. Aug 26 '11 at 1:05