Preimage of Horizontal Section of Function is Horizontal Section of Preimage

Definition
Let $X$ and $Y$ be sets.

Let $f : X \times Y \to \overline \R$ be an extended real-valued function.

Let $y \in Y$.

Let $D \subseteq \R$.

Then:


 * $\map {\paren {f^y}^{-1} } D = \paren {\map {f^{-1} } D}^y$

Proof
Note that:


 * $x \in \map {\paren {f^y}^{-1} } D$




 * $\map {f^y} x \in D$

That is:


 * $\map f {x, y} \in D$

This is equivalent to:


 * $\paren {x, y} \in \map {f^{-1} } D$

Which in turn is equivalent to:


 * $x \in \paren {\map {f^{-1} } D}^y$

So:


 * $x \in \map {\paren {f^y}^{-1} } D$ $x \in \paren {\map {f^{-1} } D}^y$.

giving:


 * $\map {\paren {f^y}^{-1} } D = \paren {\map {f^{-1} } D}^y$