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:

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

where:

$f^y$ is the $y$-horizontal section of $f$

$\paren {f^{-1} \sqbrk D}^y$ is the $y$-horizontal section of $f^{-1} \sqbrk D$.

Proof

Note that:

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

if and only if:

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

from the definition of preimage.

That is, by the definition of the $y$-horizontal section:

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

From the definition of preimage, this is equivalent to:

$\tuple {x, y} \in f^{-1} \sqbrk D$

Which in turn is equivalent to:

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

from the definition of the $y$-horizontal section.

So:

$x \in \paren {f^y}^{-1} \sqbrk D$ if and only if $x \in \paren {f^{-1} \sqbrk D}^y$.

giving:

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

$\blacksquare$