# properties of functions

Let $f\colon X\to Y$ be a function. Let $(A_{i})_{i\in I}$ be a family of subsets of $X$, and let $(B_{j})_{j\in J}$ be a family of subsets of $Y$, where $I$ and $J$ 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 $f(A)$)

• $f(f^{-1}(B))\subseteq B$ 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 $f$ is surjective.

For more properties related specifically to inverse images, see the inverse image (http://planetmath.org/InverseImage) entry.

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

• $f$ 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 $S,T\subseteq X$

Title properties of functions PropertiesOfFunctions 2013-03-22 14:59:54 2013-03-22 14:59:54 yark (2760) yark (2760) 20 yark (2760) Result msc 03E20 PropertiesOfAFunction