Union of Inverses of Mappings is Inverse of Union of Mappings
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Theorem
Let $I$ be an indexing set.
Let $\family {f_i: i \in I}$ be an indexed family of mappings.
For each $i \in I$, let $f^{-1}$ denote the inverse of $f$.
Then the inverse of the union of $\family {f_i: i \in I}$ is the union of the inverses of $f_i, i \in I$.
That is:
- $\ds \paren {\bigcup \family {f_i: i \in I} }^{-1} = \bigcup \family {f_i^{-1}: i \in I}$
Proof
\(\ds \tuple {y, x}\) | \(\in\) | \(\ds \paren {\bigcup \family {f_i: i \in I} }^{-1}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \family {f_i: i \in I}\) | Definition of Inverse Relation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists i \in I: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds f_i\) | Definition of Union of Family | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists i \in I: \, \) | \(\ds \tuple {y, x}\) | \(\in\) | \(\ds f_i^{-1}\) | Definition of Inverse Relation | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \tuple {y, x}\) | \(\in\) | \(\ds \bigcup \family {f_i^{-1}: i \in I}\) | Definition of Union of Family |
$\blacksquare$