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I'm very glad about this topic, because I have a similar problem and Lemma 5.2 by Davydov would be the solution. But I have a problem with the proof. Davydov says that the inclusion $\langle a \rangle \to A$ and the surjection $A \to \widehat{\langle a \rangle}$ split. Does that mean that the short exact sequence $$ 0 \to \langle a \rangle \to A \to \widehat{\langle a \rangle} \to 0 $$ splits? But then the Splitting Lemma provides an isomorphism $A \cong \langle a \rangle \oplus \widehat{\langle a \rangle}$, but Davydov states that $$ A \cong \langle a \rangle^{\perp}\big/\langle a \rangle \oplus \langle a \rangle \oplus \widehat{\langle a \rangle}. $$ Could anyone help me by understanding Davydovs proof or do you have an alternative source (proof) for me?

Thanks a lot!
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I'm very glad about this topic, because I have a similar problem and Lemma 5.2 by Davydov would be the solution. But I have a problem with the proof. Davydov says that the inclusion $\langle a \rangle \to A$ and the surjection $A \to \widehat{\langle a \rangle}$ split. Does that mean that the short exact shourt sequence $$ 0 \to \langle a \rangle \to A \to \widehat{\langle a \rangle} $$ splits? But then the Splitting Lemma provides an isomorphism $A \cong \langle a \rangle \oplus \widehat{\langle a \rangle}$, but Davydov states that $$ A \cong \langle a \rangle^{\perp}\big/\langle a \rangle \oplus \langle a \rangle \oplus \widehat{\langle a \rangle}. $$ Could anyone help me by understanding Davydovs proof or do you have an alternative source (proof) for me?

Thanks a lot!
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I'm very glad about this topic, because I have a similar problem and Lemma 5.2 by Davydov would be the solution. But I have a problem with the proof. Davydov says that the inclusion $\langle a \rangle \to A$ and the surjection $A \to \widehat{\langle a \rangle}$ split. Does that mean that the exact shourt sequence $$ 0 \to \langle a \rangle \to A \to \widehat{\langle a \rangle} $$ splits? But then the Splitting Lemma provides an isomorphism $A \cong \langle a \rangle \oplus \widehat{\langle a \rangle}$, but Davydov states that $$ A \cong \langle a \rangle^{\perp}\big/\langle a \rangle \oplus \langle a \rangle \oplus \widehat{\langle a \rangle}. $$ Could anyone help me by understanding Davydovs proof or do you have an alternative source (proof) for me?

Thanks a lot!