If $P$ is projective and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$

Question

I need to prove that if $P$ is a projective module over $\mathbb Z$ and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$.

This in turn would imply if $A$ and $B$ are projective then $A\cap B$ is projective. (Since $A$ and $B$ would be direct summands of $A+B$ which is clearly also projective.

Answer 1

We have an exact sequence $0\to P/A\cap B\stackrel{f}\to P/A\oplus P/B$. (Define $f(x)=(x,-x)$.) Then $P/A\cap B$ is a submodule of a projective module which over $\mathbb Z$ is free, so $P/A\cap B$ is also free.