# Wild Spectral Sequences Ep. 4: Schanuel’s 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 in working with total complexes (however, to appreciate it you just need a basic grasp of module theory or even just abelian groups). It goes like this: suppose $f:A\to B$ is a module homomorphism. What is the kernel of $A\oplus B\to B$ give by $(a,b)\mapsto f(a) + b$? By just using the definition of kernel, we see that it is $\{ (a, -f(a)) : a\in A\}$. which is actually just isomorphic to $A$ itself. This handy fact will come it with our proof of Schanuel’s lemma.

### The Proof

Let us restate the theorem.

Theorem. 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 left $R$-modules with $P$ projective, then $K_1\oplus P_2\cong K_2\oplus P_1$.
Proof. Since $P_1$ is projective, we can put the two short exact sequences into the commutative diagram:

$\begin{matrix} 0 & \to & K_1 & \to & P_1 & \to & A & \to & 0\\ ~ & ~ & \downarrow & ~ & \downarrow & ~ & \parallel & ~ & ~ \\0 & \to & K_2 & \to & P_2 & \to & A & \to & 0\end{matrix}$

Next, we observe that since the rows are exact, the spectral sequence $H_p^v(H_q^h(C)\Rightarrow H_{p+q}(\mathrm{Tot}(C))$ has zero $E^\infty$ terms so that the total complex is exact:

$0\to K_1\to K_2\oplus P_1\to P_2\oplus A\to A\to 0$.

Now we recall that the kernel of $P_2\oplus A\to A$ is just isomorphic to $P_2$ via the ‘expanding the kernel’ trick, and so the image of $K_2\oplus P_1\to P_2\oplus A$ is isomorphic to $P_2$. Hence we get a short exact sequence:

$0\to K_1\to K_2\oplus P_1\to P_2\to 0$

Since $P_2$ is projective, this sequence splits giving us the isomorphism $K_2\oplus P_1\cong K_1\oplus P_2$. QED

Notice that once we get used to spectral sequences, they can help remove a lot of the clutter that comes with ridiculous proofs that contain sentences of the form ‘let $x\in X$, then $f(x)\in Y$ is in the kernel of…’, which are exceptionally hard to read.