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I am trying to understand the particular argument in Lemma 15.102.2's proof. It writes that if $M$ is invertible $R$-module, then

we have an automorphism $M \rightarrow M$ which factors as $$M \rightarrow R^n \rightarrow M.$$

Then it says $M$ is a direct summand of $R^n$. How is this so?

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In this case, there are homomorphisms $\phi:R^n\to M$ and $\psi:M\to R^n$ with $\psi\circ\phi=\text{id}_M$. Then $R^n$ is the direct sum of $\text{im}\,\psi$ and $\ker\phi$.

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  • $\begingroup$ The problem is I don't see where $\psi$ comes from. $\endgroup$ – CL. May 6 at 11:41
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    $\begingroup$ @CL. If $u\circ v=w$ and $w$ is an automorphism, then compose with $w^{-1}$ and get... $\endgroup$ – user26857 May 6 at 18:03

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