Preimage of Vertical Section of Function is Vertical Section of Preimage

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:


 * $y \in \paren {f_x}^{-1} \sqbrk D$ $y \in \paren {f^{-1} \sqbrk D}_x$.

giving:


 * $\paren {f_x}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}_x$