Weibel IHA Exercise 1.2.5

I have started to work through 'An introduction to homological algebra' by Weibel and spend more time than I want going in circles on exercise 1.2.5.

The exercise states the following:

Proof in an elementary way (not with spectral sequences or the Acyclic assembly lemma) that if $C$ be a bounded double complex with exact rows or columns then $Tot(C)$ is acyclic.

This seems to be a basic diagram chase exercise, which is why I chased with determination and wasted hours on this small exercise. Could someone give me a hint/answer?

Solution 1:

This is an old question, but since it took me a while to solve this problem, I will tell you the idea by walking through an example. Consider the double complex \begin{array}{ccccccccccc} & & 0 & & 0 & & 0 & & 0 & & \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \leftarrow & C_{0,2} & \leftarrow & C_{1,2} & \leftarrow & C_{2,2} & \leftarrow & C_{3,2} & \leftarrow & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \leftarrow & C_{0,1} & \leftarrow & C_{1,1} & \leftarrow & C_{2,1} & \leftarrow & C_{3,1} & \leftarrow & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \leftarrow & C_{0,0} & \leftarrow & C_{1,0} & \leftarrow & C_{2,0} & \leftarrow & C_{3,0} & \leftarrow & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0 & & 0 & & 0 & & 0 & & \end{array} with exact rows. Let us look at the total complex $T(C)$, and say we concentrate on the diagonal for $n=3$, that is, we look at \begin{align} T(C)_3 = C_{1,2} \oplus C_{2,1} \oplus C_{3,0}, \end{align} and say we have some element $x+y+z\in T(C)_3$, with $x\in C_{1,2}$ etc.

We want to show: if $d(x+y+z) = 0$ then there is some $\xi \in T(C)_4$ such that $d\xi = x+y+z$.

What is $d(x+y+z)$? It is \begin{align} d(x+y+z) = d^h x + d^v x + d^h y + d^v y + d^h z + d^v z. \end{align} Note that then $d(x+y+z)=0$ is equivalent to the four conditions \begin{align} 0 &= d^h x \\ 0 &= d^v x + d^h y \\ 0 &= d^v y + d^h z \\ 0 &= d^v z, \end{align} and the first immediately implies there is some $a\in C_{2,2}$ such that $$ d^h a = x. $$ Then, we plug it into the second condition, to get $$ 0 = d^v x + d^h y = d^v d^h a + d^h y = d^h(y - d^v a) $$ so that $y - d^v a = d^h b$ for some $b\in C_{3,1}$. As you may have guessed, the third condition tells us $$ 0 = d^v y + d^h z = d^h(z - d^v b) $$ and this would continue on and on, but in this example, the map $d^h: C_{3,0} \to C_{2,0}$ is actually injective, so $z = d^v b$. In our example, the fourth condition doesn't even matter.

Now, $\xi = a + b$ satisfies $$ d\xi = d^h a + (d^v a + d^h b) + d^v b = x + y + z, $$ neato!

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So, the general strategy is: Go to the top vertical component in the diagonal. This exists since $C$ is bounded. The condition of an element of $T(C)_n$ being in the kernel of $d$ in particular implies that the bit of that element that lives in the top of the diagonal is in the kernel of $d^h$, and $d^h$ is the differential of an exact sequence. So you can pull it back, and then...

And eventually, this chain terminates: $C$ is bounded. And you're done.