If $e,f\in R$ - an associative ring with $1$, are idempotents, then $eR\cong fR$ iff there are $u,v\in R$ such that $uv=e$ and $vu=f$. I am reading Lambek's Lectures on Rings and Modules and I have a problem with a proposition in theCompletely Reducible Rings section. The proposition is as follows:
Let $R$ be an associative ring with $1$ and $e,f\in R$ are idempotents, i.e $e^2=e$ and $f^2=f$. Then $eR\cong fR$ iff there are $u,v\in R$ such that $uv=e$ and $vu=f$.
I have a problem understanding the proof of the "only if" part, that is proving the existence of such $u,v$ if $eR\cong fR$. The proof given is:
Let $u=fue$ respond to the isomorphism $eR\to fR$ and $v=evf$ respond to the inverse isomorhpism. I have trouble understanding what the author is doing. Maybe writhing $e\in fR$ and $e=fs$ for some $s\in R$ and doing the same for $f\in eR$ could do it, but I didn't manage to proove it.

I had a hard time picking up on what he was saying at first too.
The author is using Lemma 1 preceding your proposition, that $Hom(eR_R,fR_R)\cong fRe$ and $Hom(fR_R,eR_R)\cong eRf$.  Then your isomorphism $U:eR\to fR$ is equal to $fue$ for for some $u$, and $V:fR\to eR$ is equal to $evf$ for some $v$.
Because $VU$ is the identity on $eR$, $e=VUe=evffuee=(evf)(fue)$.
Likewise because $UV$ is the identity on $fR$, $f=fueevff=(fue)(evf)$.
Now you can see the two elements multiplying in opposite orders to yield $e$ and $f$.
