----- > [!proposition] Proposition. ([[preimages and unions commute]]) > Consider a function $f:X\to Y$. Let $A, B \subset Y$. Then we have $f^{-1}(A \cup B)=f^{-1}(A) \cup f^{-1}(B).$ > [!proof]- Proof. ([[preimages and unions commute]]) > Let $x \in X$ s.t. $y:=f(x)$ is in $A \cup B$, that is, $x \in f^{-1}(A \cup B)$. Then $y \in A$ or $y \in B$ and therefore $f^{-1}(y)$ is in $f^{-1}(A)$ or $f^{-1}(B)$. Hence $f^{-1}(y) = x \in f^{-1}(A) \cup f^{-1}(B).$ \ For the reverse inclusion, let $x \in f^{-1}(A) \cup f^{-1}(B)$. Then $y \in A$ or $y \in B$ and we are done. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```