Direct sums in projective modules Let $P$ be a projective module and $P=P_1+N$, where $P_1$ is a direct summand of $P$ and $N$ is a submodule. Show that there is $P_2\subseteq N$ such that $P=P_1\oplus P_2$. 
I know that there is a submodule $P'$ of $P$ such that $P=P_1\oplus P'$. I wanted to consider the projection from this to $P_1$ and use the definition of being projective. But I would also need a map from $P=P_1+N$ to $P_1$ and I don't know how to get a well defined map there because it is not a direct sum. 

The condition $P = P_1 + N$ implies that the natural map $N \to P/P_1$ is surjective.  As $P$ is projective this means the quotient map $P \to P/P_1$ factors through $N$, so there is a homomorphism $P \to N$ such that the composition $P \to N \to P/P_1$ is the quotient map.  Since the quotient map gives an isomorphism $P' \simeq P/P_1$ we can define $P_2 \subseteq N$ to be the image of $P'$ under the map $P \to N$.  Now $P_1$ and $P_2$ don't intersect because the quotient map was injective on $P'$ and $P = P_1 + P_2$ because $P'$ surjects onto $P/P_1$.  Thus $P = P_1 \oplus P_2$.
