Automorphism of $P\oplus P'$ that induces isomorphism $P\oplus K'\cong P'\oplus K$. 
Let $P,P'$ be two projective $R$-modules, and $K\subset P,K'\subset P'$ be submodules  such that $P/K\cong P'/K'$. Show that there exists an automorphism $\phi$ of $P\oplus P'$ such that $\phi(P\oplus K')=P'\oplus K$. In particular, $P\oplus K'\cong P'\oplus K$.

It is not difficult to show directly that $P\oplus K'\cong P'\oplus K$, for instance, by taking the pullback square of $P\to P/K, P'\to P'/K'\cong P/K$. But I don't know how to construct such a $"\phi"$ in the question. Any help?

Just recall the classical definition in the process of proving Schanuel's lemma
$$ X=\{ (p,p')\in P\oplus P' :\pi(p)=\pi '(p') \} \; .$$
And try to write down explicit maps how $X$ isomorphic to $P\oplus K'$ and $K\oplus P'$  using two lifted maps $P\rightarrow P'$ and $P'\rightarrow P$ compatible with the projection.
