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Let $(A_\nu)_{\nu<\mu}$ be an increasing sequence of free abelian groups such that for any limit ordinal $\lambda<\mu$, $A_\lambda=\bigcup_{\nu<\lambda}A_\nu$. Let $A=\bigcup_{\nu<\mu}A_\nu$. Let $\kappa$ be the cofinality of $\mu$.

Assume that $A_\sigma/A_{\nu+1}$ is free whenever $\nu<\sigma<\mu$.
Let $\{p_\sigma:\sigma<\kappa\}$ be a closed unbounded subset of $\mu$ that is strictly increasing. Assume that for every $\sigma<\kappa$ such that $p_\sigma$ is a limit ordinal, $A_{p_\sigma+1}/A_{p_\sigma}$ is free.

A paper says it is "evident" that $A$ is free and that $A/A_{p_\sigma}$ is free for all $\sigma<\kappa$.

It is not evident to me. Why are these statements true?

Source: the proof of Lemma 2.1 in Paul Eklof, "On the Existence of $\kappa$-Free Abelian Groups," Proceedings of the American Mathematical Society 47 (1975), 65-72.

Tri
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  • I'm not sure if this is what you're looking for, but in Rotman's Homological Algebra, on Page 103, just after Example 3.7, he proves that if $P=\cup_\alpha P_\alpha$, and if each $P_\alpha$ is a direct summand of each $P_{\alpha+1}$, and $P_\alpha=\cup_{\beta<\alpha} P_\beta$ for every limit ordinal $\alpha$, then $P\cong\oplus_{\alpha} P_{\alpha+1}/P_\alpha$. – Ashwin Trisal Jul 15 '18 at 02:52
  • Thank you. That was the key. – Tri Jul 15 '18 at 05:42

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