For modules one has the isomorphism theorem $(A/C)/(B/C) \cong A/B$ for $C\leq B\leq A$. One way to remember it is through analogy with canceling of fractions. Another way to remember and prove it is to put all the modules in a 3×3 commutative diagram

$$

\begin{matrix}

C & \to & B & \to & B/C\\

\downarrow & ~ & \downarrow & ~ & \downarrow\\

C & \to & A & \to & A/C\\

\downarrow & ~ & \downarrow & ~ & \downarrow\\

0 & \to & A/B & \to & (A/C)/(B/C)

\end{matrix}

$$

where there are also zero arrows on the edges of the diagram, but I have omitted them for ease of typesetting. All the columns are exact, and the first two rows are exact, so the remaining row is exact giving the required isomorphism via the 3×3-lemma.

It's time for another installment of Wild Spectral Sequences! We shall start our investigations with a classic theorem useful in many applications of homological algebra called Schanuel's lemma, named after Stephen Hoel Schanuel who first proved it.

Consider for a ring $ R$ the category of left $ R$-modules, and let $ A$ be any $ R$-module. Schanuel's lemma states: if $ 0\to K_1\to P_1\to A\to 0$ and $ 0\to K_2\to P_2\to A\to 0$ are exact sequences of $ R$-modules with $ P$ projective, then $ K_1\oplus P_2\cong K_2\oplus P_1$.

We shall prove this using spectral sequences. I came up with this proof while trying to remember the "usual" proof of Schanuel's lemma and I thought that this would be a good illustration of how spectral sequences can be used to eliminate the dearth of clarity in the dangerous world of diagram chasing.

Before I start, I'd like to review a pretty cool fact I which I think of as *expanding the kernel*, which is pretty useful More »