Projective module which is direct sum of $R^n$

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Proposition 4.7.5 Let $P$ be a left $R$-module and $\{u_i\}$ be a generating set for $P$.

Then $P$ is projective if and only if there exists $R$-module homomorphisms $\alpha_i:P\rightarrow R$ such that

(1) $\alpha_i$ is zero on all except finitely many $u_i$'s.

(2) Every $x\in P$ can be written as $x=\sum_i \alpha_i(x)u_i.$

This is from Cohn's Basic Algebra. Then there is a corollary

The case when the family $\{u_i\}$ is finite is worth stating separately:

Corollary 4.7.6 The module $P$ is a direct sum of $R^n$ if and only if there exists $u_1,u_2,\ldots, u_n\in P$ and $\alpha_1,\cdots,\alpha_n\in {\rm Hom}_R(P,R)$ such that equation in (2) holds.

Q. I did not understand statement of the corollary. $P$ is direct sum of $R^n$ means $R^n\cong P\oplus Q$ for some $R$-module $Q$? How this corollary follows from proposition?

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Sure it is correct. We can consider the maps

$\alpha : R^n \to P$ such that $\alpha(a_1,\dots ,a_n):=\sum_{k=1}^n a_k u_k$

and

$\beta : \ker(\alpha) \to R^n$

the inclusion map. Then

$0\to \ker(\alpha) \to R^n \to P\to 0$

it is exact but P is projective so the sequence split:

$R^n=P\oplus \ker(\alpha)$