There is a well-known property of the vector spaces: for every vector $v \in V$ there exists (maybe, it isn't unique) basis $\mathcal B \subset V$ such that $v \in \mathcal B$.
It means that for every element $w$ of the direct sum (maybe, infinite) $V^{\oplus I} := V \oplus V \oplus V$ ... $\oplus V \oplus$ ... ($I$ is some set of indexes) there is a morphism $\phi: V \to V^{\oplus I}$ such that $w \in \operatorname{Im}\phi$.
I want to use this property for the modules (maybe, over the non-commutative ring) but I don't know, true or false is it...
My question:
Let $P$ be a projective module over a ring $R$. Is it true that for every element $x$ of the direct sum (maybe, infinite) $P^{\oplus I} := P \oplus P \oplus P$ ... $\oplus P \oplus$ ... ($I$ is some set of indexes) there is a morphism $\phi: P \to P^{\oplus I}$ such that $x \in \operatorname{Im}\phi$?
UPD: As Eric Wofsey said, it's enough to prove this fact only for $I = {1, 2}$: the simple induction'll finish our work for finite $I$'s. The case of infinite $I$ is also simple: every element of $P^{\oplus I}$ is contained in some finite direct sum, so the previous part of argument works. But what to do if $|I|=2$? I don't know...
No, this is not true. For instance, let $R$ be an integral domain that has a nonprincipal invertible ideal $P\subset R$ generated by two elements $a$ and $b$ (e.g., $R=\mathbb{Z}[\sqrt{-5}]$, $P=(2,1+\sqrt{-5})$). I then claim that the element $(a,b)\in P^2$ is not in the image of any homomorphism $f:P\to P^2$. Indeed, every homomorphism $P\to P$ is given by multiplication by some element of $R$, and so there must be elements $r,s\in R$ such that $f(x)=(rx,sx)$ for all $x\in P$. If $f(x)=(a,b)$, then this means $a$ and $b$ are both multiples of $x$ which then means that $x$ generates $P$. Since $P$ is nonprincipal, this is a contradiction.