# $P$ direct summand of a free module implies $P$ is projective

I've looked at several posts here and proofs elsewhere and it is not clicking in my brain.

Let $$R$$ a ring and $$P$$ be a direct summand of a free module. That is, there exists a (left) $$R$$-module $$Q$$ such that $$F=P \oplus Q$$ is free. Then, let $$B= \{b_i\}$$ be a basis for $$F$$. Further suppose we have an epimorphism $$f:A \to B$$ and a homomorphism $$g:P \to B$$.

Here is the bit that confuses me. Since $$g$$ is an epimorphism we can find $$a_i \in A$$ such that $$f(a_i)=g(b_i)$$ for each $$i$$. $$g$$ is only a map from $$P \to B$$ not from $$F \to B$$. How are we able to make a consideration like this? Why does it work out?

From there, define $$\bar{h}:F \to A$$

$$\bar{h}(\sum r_ib_i)=\sum r_im_i$$

We can check that this map is well-defined and what not and restricting to $$P$$ gives the desired result.

You can use that the Hom functor commutes with direct sums: we have a canonical isomorphism $$\DeclareMathOperator{\Hom}{Hom} \Hom_R(P,X)\oplus\Hom_R(Q,X)\simeq \Hom_R(F,X),$$and we can associate to the homomorphism $$g:P\to B$$ the homomorphism $$h=g\oplus 0:P\oplus Q\to B$$.
• If they belong to $Q$, yes. Their $P$-componet will also have an image, possibly non-zero. But I don't think you have to worry with a basis of $F$. Oct 25, 2018 at 18:47