Projective R-modules.

Question

We say that an $R-$module $P$ is projective if for all surjectives $R-$linear maps $g:M \to N$ and any $R-$linear map $f:P \to N$ there exist an $R-$linear map $h:P \to M$ such that $g \circ h=f$.

I am trying to prove the following claim:

Every projective module $P$ is a direct summand of some free module $F$, that is, $F=P \oplus Q$ for some $Q$.

What did I do so far: I started taking $F$ as being the free $R-$module over $P$, that is, $F$ is such that $P$ is (or contains) a basis for $F$ (To be honest I am not 100% sure that I can do that, so if someone can, please tell me if this argment is valid). Then there is a natural surgection $g: F \to P$ by mapping the basis elements of $F$ into itself in $P$. And, we have the identity map $i:P \to P$, and then since $P$ is projective, there exist a map $h: P \to F$ such that $g \circ h = i$.

Okay, now since $g \circ h=i$ it is not difficult to show that $F = Ker(g) \oplus Im(h)$. Indeed given $x \in F$ we have $g(x) \in Im(g)=P$, by surjection, then $h(g(x))$ is well defined and belongs to $F$, then we can set $x=(x-h(g(x)))+h(g(x)) \in Ker(g)+Im(h)$ since $g(x-h(g(x)))=g(x)-g(h(g(x)))=g(x)-g(x)=0$.

Moreover $Ker(g) \cap Im(h)=\{0\}$ since if $x \in Ker(g) \cap Im(h)$ then $x=h(a)$ for some $a \in P$ then $0=g(x)=g(h(a))=a$, then $x=0.$ Then $$F=Ker(g) \oplus Im(h),$$ however I cannot see why one of them is equal to $P$. Can someone give me some hint to go to the next step?? I have tried to show that $F = Im(g) \oplus Ker(h)$ but I could not, I think it does not make sense!

Thank you very much, people! Have a good saturday!

Answer 1

Since $g\circ h=\operatorname{id}_P$, $h$ is injective, so that $\;P\simeq\operatorname{Im h}$ by the first isomorphism theorem.

Answer 2

You can exchange the passage " I started taking $F$ as being the free $R$−module over $P$, that is, $F$ is such that $P$ is (or contains) a basis for $F$" by: since every $R$-module is quotient of an free $R$-module, let $L$ be a set such that the canonical map $R^{(L)}\longrightarrow P$ is surjective (we can take $L=P$, for example). So we have an exact sequence $$0\longrightarrow K\longrightarrow R^{(L)}\longrightarrow P\longrightarrow0$$ where $K$ is the kernel of the canonical surjection.

From here use the hypothesis for conclude that the canonical surjection admit a section and therefore $R^{(L)}\simeq P\oplus K$.