Below is part of the the question that i have to prove.
f is injective then $f•f^{-1}(y)=y$ for all $y\in R(f)$ where $R(f)$ is range of $f$
My attempt: let $f$ be injective and $y=f(x)$ then $f^{-1}(y)=x$, so that as $f$ injective above gives, $f(f^{-1}(y))=f(x)=y$ and hence result. Am i correct?
(Is there is any detail that i left) or it is sufficient.
Your idea is correct. But in an exam I would not give you the full points. You did not even mention the range of $f$ in your proof although it is an essential part of the question.
Let $y\in R(f)$. Then there is a $x\in D(f)$ such that $f(x) = y$. Due to injectivity of $f$ this $x$ is unique. Thus $f^{-1}(y) = x$ and $f(f^{-1}(y)) = f(x) = y$.