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Let $\{G_i, \phi_{i,j}, i, j \in I\}$ (with a partially ordered $I$) be an inverse system of finite groups and let $G = \varprojlim G_i$ be its inverse limit, that is, a profinite group. If I know that all $\phi_{i,j}$ are surjective how can I conclude that the canonical projections $\phi_i: G \to G_i$ are surjective too?

My idea: If I have $x_i \in G_i$ then I can define the components of a $y \in G$ in following way:

For $i$: $y_i := x_i$; for j with $i>j$: $y_j := \phi_{i,j}(x_i)$. If $I$ is countable I can for every $k>i$ pull back the components $y_k$ of $y$ in following way by induction: $\phi_{k,i}$ is surjective so I can find a $x_k \in G_k$ with $\phi_{k,i}(x_k) = X_i$. So define $y_k:= x_k$. Because is a inverse system we get in this way an $y \in G$. But how can I pull back the components $y_k$ of y for $k>i$ to get a $y \in G$ with $\phi_i(y)= X_i$ if $I$ isn't countable?

Sampah
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user267839
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1 Answers1

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Note that the $G_i$ are all finite, this means if we equip the $G_i$ with the discrete topology, all subset of $G_i$ are also compact. Now for every $\ell\in I$, we take the the following sets $X_\ell\subset G_\ell$: For $i\geq\ell$ we have $X_\ell=\{\phi_{i,\ell}(x_i)\}$ For $i\leq\ell$ we have $X_\ell=\{\phi_{\ell,i}^{-1}(x_i)\}$. Because $\phi_{\ell,i}$ are surjective we have $X_\ell\neq\emptyset$ for all $\ell\in I$ and the $X_\ell$ are all compact. Moreover $\phi_{\ell,k}(X_\ell)\subset X_k$ holds for all $\ell\geq k$. Hence $\{X_\ell,\phi_{\ell,k},\ell,k\in I\}$ is an inverse system. Because the $X_\ell$ are all compact we have $X=\varprojlim X_\ell\neq\emptyset$ and $y\in X\subset G$ satisfies $\phi_i(y)=x_i$.

Sniper
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