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Consider a function $h: \mathbb{R}^2\to \mathbb{R}$. I define that "$h$ has an inverse in the second component" if there exists a function $h^{-1}: \mathbb{R}^2\to \mathbb{R}$ such that for all $x,y\in \mathbb{R}$ holds $$y= h^{-1}(h(x,y),x).$$ In other words, if $h(x,y)=z$ then $h^{-1}$ is a function such that $y=h^{-1}(z,x)$. Denote a family $I=\{h: h \text{ has an inverse in the second component}\}$.

Can you tell me something about this family $I$? E.g. if $h$ is smooth, does it imply that $h\in I$? Probably not, but I would like to have some sufficient condition that is easily checked.

Albert Paradek
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    Inverse function theorem. If $h$ is differentiable and the partial derivative with respect to $y$ of $h(x,y)$ is invertible then $h$ is in $I$. This holds even for locally Lipschitz functions (See the book of Clarke on nonsmooth optimization). – Jürgen Sukumaran Mar 29 '22 at 15:23
  • A continuous function of a single variable is invertible if and only if it is strictly increasing or strictly decreasing. The same holds for $y\mapsto h(x,y)$ when that's continuous for every fixed $x$. I believe you want to write $y=h^{-1}(x,h(x,y))$. The argument $x$ is just a "bystander". You can apply everything we know about inverse functions to the one- variable function $y\mapsto h(x,y)$. – Kurt G. Mar 29 '22 at 18:47

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