how to prove inverse of a bijective right linear function is right linear?

Question

Let $R$ be a ring, $A$ and $B$ be right $R$-modules. A right $R$-linear function $f:A \rightarrow B$ is defined as $f(xa+yb)=f(x)a+f(y)b$. If $f$ is $1-1$, onto and right linear, then how to prove that $f^{-1}$ is right linear?

Answer 1

Let $u, v \in B$, and $a,b \in R$.

First, there exists $x,y\in A$ such that $u = f(x)$ and $v=f(y)$

Then,

$$f^{-1}(ua+vb) = f^{-1}(f(x)a+f(y)b)$$

Now, by right R-lineary of $f$, it's equal to

$$= f^{-1}(f(xa+yb) ) = xa+yb$$

And finally, as $x = f^{-1}(u)$ and $y = f^{-1}(v)$

$$ = f^{-1}(u)a+f^{-1}(v)b$$