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Let $(I, \le)$ be a poset and $\{ A_i \}$ a collection of sets indexed by $I$, such that the projections $\varprojlim A_i \to A_i$ are surjective for all $i \in I$.

Is it true that the natural map $$\varprojlim_{i \in I} A_i \to \varprojlim_{j \in J} A_j$$ is surjective whenever $J \subseteq I$?

This feels true: given a coherent sequence $(a_j)_{j \in J}$, I should be able to use surjectivity of all maps $A_i \to A_{i'}$ to "connect the dots" and obtain a suitable preimage sequence $(a_i)_{i \in I}$. Two failed attempts at a proof have been to use Zorn's lemma on the non-empty posets $$\Sigma = \{ K \subseteq J: \varprojlim_{i \in I} A_i \twoheadrightarrow \varprojlim_{i \in K} A_i \},$$ $$\Sigma' = \{ I \supseteq K \supseteq J: \varprojlim_{i \in K} A_i \twoheadrightarrow \varprojlim_{i \in j} A_i \}.$$ A proof or a counter-example would be appreciated.

Asaf Karagila
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Mr. Chip
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1 Answers1

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Turns out this is wrong. Let $A = \{ 0, 1 \}$ and $A_n = A^n$ for any $n \in \mathbb{N}$, with connecting maps $\pi_n: A_n \to A_{n-1}$ given by projection onto the first $n - 1$ coordinates. Also let $A_\infty$ denote the set of eventually zero sequences in $A^\mathbb{N}$.

Now if $I = \mathbb{N} \cup \{ \infty \}$ and $J = \mathbb{N}$, one finds that $\varprojlim_{i \in I} A_i = A_\infty$, while $\varprojlim_{j \in J} A_j = A^\mathbb{N}$, so the natural map described in the question is not surjective.

Mr. Chip
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