If $f: A \to B$ is given by $f(x,y) = (x,y,0)$, where $A$ is an open disc in $\mathbb{R^2}$ and $B=\{{(x,y,0)}\in \mathbb{R^3} \mid (x,y) \in A\}$ what is its inverse frunction $f^{-1}$?
2026-03-28 22:13:46.1774736026
What is the inverse function of $f(x,y) = (x,y,0)$?
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The inverse is $f^{-1}(x,\,y,\,0)=(x,\,y)$.