Lemma 52.3.2. Suppose $X$, $f$, $(\mathcal{F}_ n)$ is as in Lemma 52.3.1. Then the limit topology on $H^ p = \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}_ n)$ is the $f$-adic topology.

Proof. Namely, it is clear that $f^ t H^ p$ maps to zero in $H^ p(X, \mathcal{F}_ t)$. On the other hand, let $c \geq 1$. If $\xi = (\xi _ n) \in H^ p$ is small in the limit topology, then $\xi _ c = 0$, and hence $\xi _ n$ maps to zero in $H^ p(X, \mathcal{F}_ c)$ for $n \geq c$. Consider the inverse system of short exact sequences

$0 \to \mathcal{F}_{n - c} \xrightarrow {f^ c} \mathcal{F}_ n \to \mathcal{F}_ c \to 0$

and the corresponding inverse system of long exact cohomology sequences

$H^{p - 1}(X, \mathcal{F}_ c) \to H^ p(X, \mathcal{F}_{n - c}) \to H^ p(X, \mathcal{F}_ n) \to H^ p(X, \mathcal{F}_ c)$

Since the term $H^{p - 1}(X, \mathcal{F}_ c)$ is independent of $n$ we can choose a compatible sequence of elements $\xi '_ n \in H^1(X, \mathcal{F}_{n - c})$ lifting $\xi _ n$. Setting $\xi ' = (\xi '_ n)$ we see that $\xi = f^{c + 1} \xi '$. This even shows that $f^ c H^ p = \mathop{\mathrm{Ker}}(H^ p \to H^ p(X, \mathcal{F}_ c))$ on the nose. $\square$

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