The Stacks project

Proof. By transfinite induction we prove this holds for every ordinal $\alpha $ and every system as in the lemma. In particular, whilst proving the result for $\alpha $ we may assume the complexes $\mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K^ n_\gamma $ are acyclic.

Let $x \in \mathop{\mathrm{lim}}\nolimits _{\beta < \alpha } K^0_\alpha $ with $\text{d}(x) = 0$. We will find a $y \in K^{-1}_\alpha $ with $\text{d}(y) = x$. Write $x = (x_\beta )$ where $x_\beta \in K_\beta ^0$ is the image of $x$ for $\beta < \alpha $. We will construct $y = (y_\beta )$ by transfinite recursion.

For $\beta = 0$ let $y_0 \in K_0^{-1}$ be any element with $\text{d}(y_0) = x_0$.

For $\beta = \gamma + 1$ a successor, we have to find an element $y_\beta $ which maps both to $y_\gamma $ by the transition map $f : K^\bullet _\beta \to K^\bullet _\gamma $ and to $x_\beta $ under the differential. As a first approximation we choose $y'_\beta $ with $\text{d}(y'_\beta ) = x_\beta $. Then the difference $y_\gamma - f(y'_\beta )$ is in the kernel of the differential, hence equal to $\text{d}(z_\gamma )$ for some $z_\gamma \in K^{-2}_\gamma $. By assumption, the map $f^{-2} : K^{-2}_\beta \to K^{-2}_\gamma $ is surjective. Hence we write $z_\gamma = f(z_\beta )$ and change $y'_\beta $ into $y_\beta = y'_\beta + \text{d}(z_\beta )$ which works.

If $\beta $ is a limit ordinal, then we have the element $(y_\gamma )_{\gamma < \beta }$ in $\mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K^{-1}_\gamma $ whose differential is the image of $x_\beta $. Thus we can argue in exactly the same manner as above using the termwise surjective map of complexes $f : K_\beta ^\bullet \to \mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K_\gamma ^\bullet $ and the fact (see first paragraph of proof) that we may assume $\mathop{\mathrm{lim}}\nolimits _{\gamma < \beta } K_\gamma ^\bullet $ is acyclic by induction. $\square$