Question about proof that every f.g. projective module over a local ring is free. I'm reading the proof here.
I'm at the line where they say
$$
\psi\pi(f)=\psi(f+FR)=\varphi(f)+PR.$$
Since $\psi\pi$ is surjective, it should follow that $\{\varphi(f)+PR:f\in F\}=P/PR$. I don't understand the notation $\mathrm{im}(\varphi)+PR=P$. I assume $\mathrm{im}(\varphi)+PR=\{\varphi(f)+PR:f\in F\}$? If that's a subset of $P/PR$, how can it equal $P$? 

(This should just be a comment, but too many of these questions are left without posted answers...)
The notation $\text{im}(\varphi) + PR$ refers to the sum of the two submodules $\text{im}(\varphi)$ and $PR$ (these are subsets of $P$, not $P/PR$). Explicitly,
$$\text{im}(\varphi) + PR = \{\varphi(f) + \sum_{\text{finite}} pr \mid f \in F, p \in P, r \in R \}$$
The reason that $\text{im}(\varphi) + PR = P$, is the equality $\{\varphi(f)+PR:f\in F\}=P/PR$, which says that $\{\varphi(f)\}$ forms a complete set of coset representatives of $PR$ in $P$, so every element of $P$ is a sum of some $\varphi(f)$ with an element of $PR$. 
