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:
- Is this proof correct and complete?
- 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.