Extensions which define the same element of $\text{Ext}^n(M,N)$ are in fact equivalent

It is well known (and wouldn't be so-named unless it were) that:

If $\xi$, $\eta$ are $n$-fold extensions of $N$ by $M$ (modules over a ring $R$) which yield the same element of $\text{Ext}^n(M,N)$, then they are in fact equivalent.

I am trying to understand a certain proof of this fact (and indeed the proof, for me, is as important as the result). My question involves a number of commutative diagrams, so I have decided to post it as a Latex pdf file. You can find it at:

http://homepages.utoledo.edu/mcrumle/mathoverflow%20letter.pdf

The proof is textbook (literally), and I am sure that I am simply having a brain fart over the thing, but it has been torturing me for some time. I'm hoping that someone can lead me out of the forest.

Thanks in advance for any help.

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"My question concerns the proof given by Benson in [1] on page 36" - I didn't found this on page 36. –  Boris Novikov Nov 6 '11 at 11:13
@Boris: It's on page 39 in the 2nd edition. –  Ralph Nov 6 '11 at 13:21

Say the maps in your first displayed diagram are, left to right between the first two rows, $\psi, f_1, f_0$ and between the second two rows $\phi, g_1, g_0$. Let $\sigma \partial_2 = \phi-\psi$ (I'm afraid my maps compose in the opposite direction to yours). Let the map $N \to Y_1$ be $\iota_Y$. You ask about the maps in the displayed equation following "Benson simply writes the diagram". Let's replace the middle $N$ by $P_2/\ker \phi$: then the maps between the first two rows are $\bar{\phi}$, $$p_1+\partial_2\ker\phi \mapsto f_1p_1 + \iota_Y \sigma p_1$$ (this turns out to be well-defined), $f_0$ then the identity. By $\bar{\phi}$ I mean the isomorphism induced by $\phi$ from $P_2/\ker\phi \to N$. Between the second two rows they are $\bar{\phi}, \bar{g_1}, g_0$, and the identity. Note that we don't need to assume $\ker \phi = \ker \psi$ or that $P_1/\partial_2\ker\phi = P_1/\partial_2 \ker \psi$.

Say two exact sequences beginning with $N$ and ending with $M$ are linked if there is a chain map between them acting as the identity on $M$ and $N$. Two such sequences represent the same element of $\operatorname{Ext}^2$ if and only if they are related by the equivalence relation generated by linkage. They need not be linked themselves, unlike for $\operatorname{Ext}^1$. So I am not sure a map between the last two sequences you write down will exist in general.

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Great answer, much appreciated. –  Mike Crumley Nov 10 '11 at 6:01

To my opinion Benson's proof is correct and the commutative diagramm in question (the second on page 2 in the pdf) always exists.

Consider the following commutative diagramm (it's the same as in your pdf but with named arrows):

$$P_2\quad \xrightarrow{D_2} \quad P_1 \quad \xrightarrow{D_1}\quad P_0$$ $$\hspace{2pt} \psi \downarrow \hspace{17pt} f_1 \downarrow \hspace{23pt} \downarrow f_2$$ $$N \quad \xrightarrow[d_2]{} \quad Y_1 \quad \xrightarrow[d_1]{} \quad Y_0$$

Futhermore let $h: P_1 \to N$ such that $\psi-\phi=h\circ D_2$, let $Z := P_1/D_2(\text{Ker } \phi)$ and let $\hat{f}_1 := f_1 -d_2\circ h: P_1 \to Y_1$.

Then on $\text{Ker }\phi$ holds: $$\hat{f}_1\circ D_2 = f_1D_2 -d_2hD_2 = f_1D_2 - d_2\psi = 0.$$ Thus $\hat{f}_1$ defines a map $\bar{f}_1: Z \to Y_1$ such that the right square in the following diagramm commutes:

$$N\quad \xrightarrow{i} \quad Z \quad \xrightarrow{D_1}\quad P_0$$ $$\hspace{35pt} id \downarrow \hspace{17pt} \bar{f}_1 \downarrow \hspace{25pt} \downarrow f_2\hspace{30pt}(\ast)$$ $$N \quad \xrightarrow[d_2]{} \quad Y_1 \quad \xrightarrow[d_1]{} \quad Y_0$$

Note that $i: N = \text{im }\phi \to Z$ is given by $i(n)=D_2(x) + \text{Ker } \phi$, if $x \in P_2$ such that $\phi(x)=n$. Hence the left-hand square also commutes.

But $(\ast)$ was just the commutative diagramm you were looking for.

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