Lemma 59.81.3. Let $R$ be an $A$-algebra which is absolutely integrally closed. Then $G(R) \to H(R)$ is surjective.
Proof. Let $h \in H(R)$ correspond to the $A$-algebra map $A[t, \frac{1}{\pi ^\ell t + 1}] \to R$ sending $t$ to $a \in A$. Since $\Phi (s)$ is monic we can find $b \in A$ with $\Phi (b) = a$. By Lemma 59.81.2 sending $s$ to $b$ we obtain a unique $A$-algebra map $A[s, \frac{1}{\pi s + 1}] \to R$ compatible with the map $A[t, \frac{1}{\pi ^\ell t + 1}] \to R$ above. This in turn corresponds to an element $g \in G(R)$ mapping to $h \in H(R)$. $\square$
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