Preimage of Horizontal Section of Function is Horizontal Section of Preimage
Jump to navigation
Jump to search
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:
- $x \in \paren {f^y}^{-1} \sqbrk D$
- $\map {f^y} x \in D$
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:
- $\paren {f^y}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}^y$
$\blacksquare$