Preimage of Vertical Section of Function is Vertical Section of Preimage
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Theorem
Let $X$ and $Y$ be sets.
Let $f : X \times Y \to \overline \R$ be an extended real-valued function.
Let $x \in X$.
Let $D \subseteq \R$.
Then:
- $\paren {f_x}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}_x$
where:
- $f_x$ is the $x$-vertical section of $f$
- $\paren {f^{-1} \sqbrk D}_x$ is the $x$-vertical section of $f^{-1} \sqbrk D$.
Proof
Note that:
- $y \in \paren {f_x}^{-1} \sqbrk D$
- $\map {f_x} y \in D$
from the definition of preimage.
That is, by the definition of the $x$-vertical section:
- $\map f {x, y} \in D$
From the definition of preimage, this is equivalent to:
- $\paren {x, y} \in f^{-1} \sqbrk D$
Which in turn is equivalent to:
- $y \in \paren {f^{-1} \sqbrk D}_x$
from the definition of the $x$-vertical section.
So:
- $y \in \paren {f_x}^{-1} \sqbrk D$ if and only if $y \in \paren {f^{-1} \sqbrk D}_x$.
giving:
- $\paren {f_x}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}_x$
$\blacksquare$