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In the proof of proposition 2.1. of Analytic.pdf there is the following map: Let $S = \lim_i S_i$ a profinite set. Let $p_i: S \rightarrow S_i$ be the projection. We can define the following map using the universal property of the limit and of the free abelian group: \begin{align*} \iota_T: \mathbb{Z}[\mathrm{Cont}(T,S)] &\longrightarrow \lim_i \underline{\mathbb{Z}[S_i]}(T) =\lim_i \mathrm{Cont}(T,\mathbb{Z}[S_i]) \\ \sum_{k \leq N} n_k [f_k] &\longmapsto \left( t \mapsto \sum_{k \leq N} n_k [p_i(f_k(t))] \right)_i \end{align*} In the proof professors Scholze and Clausen claim that the first step is to check that this map is injective for $T = *$ a point, but I could not prove it. What I managed to do is to show that for $N = 1,2,3,4,5$ if $\sum_{k \leq N} n_k [f_k]$ is in the kernel, then $n_k = 0$. But with my method is necessary to separate in a lot of cases and for arbitrary $N$ it gets messy.

How to prove that $\iota_*$ is injective?

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    $\begingroup$ Take a nonzero element $\sum_{k\leq N} n_k [f_k]$; you can assume that all $f_k\in S$ are different and all $n_k\neq 0$. Take a quotient $S\to S_i$ such that the $p_i(f_k)\in S_i$ are still different. Then the image of this element in $\mathbb Z[S_i]$ is still nonzero. $\endgroup$
    – Peter Scholze
    Commented Oct 11, 2022 at 18:03
  • $\begingroup$ @PeterScholze Thanks for the answer! Do you mean that exist $i$ such that $p_i(f_0)$, ..., $p_i(f_N)$ are all different? But what if $S = \prod_{n \in \mathbb{N}} \{0,1\}$ with $S_i = \{0, 1 \}$ and we take $f_k = (0,0,0,...,0,1,0,...)$ with $1$ in the $k$th coordinate? $\endgroup$
    – Luiz Felipe Garcia
    Commented Oct 11, 2022 at 18:17
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    $\begingroup$ Not quite a duplicate, but this question is closely related (and my answer there contains some more details). To answer the question from your comment: the term "inverse limit" usually means "limit over a [small] cofiltered index category". For each pair $i \neq j$ there exists an index $a_{i,j}$ such that $p_{a_{i,j}}(f_i) \neq p_{a_{i,j}}(f_j) \in S_{a_{i,j}}$, and since the index category is cofiltered there exists an index $a$ mapping to all $a_{i,j}$, so that $p_a(f_i) \neq p_a(f_j) \in S_a$ for all $i \neq j$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 11, 2022 at 19:16
  • $\begingroup$ @R.vanDobbendeBruyn Thank you very much! $\endgroup$
    – Luiz Felipe Garcia
    Commented Oct 11, 2022 at 19:17

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