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. 
 A: Since $P_1$ is a direct summand, there exists $P_2$ such that $P=P_1\oplus P_2$.
Consider $f:P\rightarrow P_2$ such that for every $x\in P$, write $x=x_1+x_2, x_1\in P_1,x_2\in P_2$, $f(x)=x_2$. Let $x\in P_2$, we can write $x=x_1'+n, x_1'\in P_1, n\in N$, we have $x=f(x)=f(x_1'+n)=f(n)$. This implies that the restriction $g$ of $f$ to $N$ is surjective, there exists $h:P\rightarrow N$ such that $f=g\circ h$; $P=P_1\oplus h(P_2)$.
Let $x\in P$, write $x=x_1+x_2, x_1\in P_1, x_2\in P_2$, $f(x-h(x_2))=f(x)-f(h(x_2))=x_2-x_2=0$  since $x_2=f(x_2)=f(h(x_2))$. This implies that $x-h(x_2)\in P_1$, we can write $x=x-h(x_2)+h(x_2)$ and we deduce that $P=P_1+h(P_2)$.
Let $x\in P_1\cap h(P_2)$, we can write $x=h(x_2), x_2\in P_2$, $f(x)=f(h(x_2))=f(x_2)=x_2=0$ because $x\in P_1$. This implies that $x_2=x=0$.
A: 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 $P = P_1 \oplus P_2$.
