$S=\{(x_1,x_2)\ |\ \forall x\in\mathbb{R}\colon x_1=x,x_2=\exp(x)\}$ be a set, and
$f$ be the function $f\colon S\to \mathbb{R},(x1,x2)\mapsto x_1+x_2$.
How can I show if $f$ is bijective?
How can the inverse of $f$ be calculated?
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I already know that $S\neq\emptyset$, and $S$ is an open subset of $\mathbb{R}$. Together with dimension invariance theorem follows that $f$ is bijective and has an inverse therefore.
In multivariate analysis, inverses of n-vector-valued n-vector functions are calculated. But it seems we have now a number-valued vector function.