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?
how to prove inverse of a bijective right linear function is right linear?
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linear-transformations
1 Answers
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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 $y = f^{-1}(v)$
$$ = f^{-1}(u)a+f^{-1}(v)b$$