PlanetMath: properties of functions

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properties of functions

(Result)

Let $f \colon X \to Y$ be a function. Let $(A_i)_{i \in I}$ be a family of subsets of , and let $(B_j)_{j \in J}$ be a family of subsets of , where and are non-empty index sets.

Then, it is easy to prove, directly from definitions, that the following hold:

$f(\bigcup \limits_{i \in I}{A_i}) = \bigcup \limits_{i \in I}{f(A_i)}$ (i.e., the image of a union is the union of the images)
$f(\bigcap \limits_{i \in I}{A_i}) \subseteq \bigcap \limits_{i \in I}{f(A_i)}$ (i.e., the image of an intersection is contained in the intersection of the images)
$A \subseteq f^{-1}(f(A))$ for any $A \subseteq X$ (where $f^{-1}(f(A))$ is the inverse image of )
$f(f^{-1}(B)) \subseteq B$ for any $B \subseteq Y$
$f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)$ for any $B \subseteq Y$
$f^{-1}(\bigcup \limits_{j \in J}{B_j}) = \bigcup \limits_{j \in J}{f^{-1}(B_j)}$ (the inverse image of a union is the union of the inverse images)
$f^{-1}(\bigcap \limits_{j \in J}{B_j}) = \bigcap \limits_{j \in J}{f^{-1}(B_j)}$ (the inverse image of an intersection is the intersection of the inverse images)
$f(f^{-1}(B)) = B$ for every $B \subseteq Y$ if and only if is surjective.

For more properties related specifically to inverse images, see the inverse image entry.

Further, the following conditions are equivalent (for more, see the entry on injective functions):

is injective
$f(S \cap T) = f(S) \cap f(T)$ for all $S, T \subseteq X$
$f^{-1}(f(S)) = S$ for all $S \subseteq X$
$f(S) \cap f(T) = \varnothing$ for all $S,T \subseteq X$ such that $S \cap T = \varnothing$
$f(S \setminus T) = f(S) \setminus f(T)$ for all $S,T \subseteq X$