properties of functions

Let $f:X\to Y$ 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:

• $f(\bigcup _{i\in I}{A}_i)=\bigcup _{i\in I}f(A_i)$ (i.e., the image of a union is the union of the images)

• $f(\bigcap _{i\in I}{A}_i)\subseteq \bigcap _{i\in I}f(A_i)$ (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 $B\subseteq Y$

• $f^{-1}(\bigcup _{j\in J}{B}_j)=\bigcup _{j\in J}{f}^{-1}(B_j)$ (the inverse image of a union is the union of the inverse images)

• $f^{-1}(\bigcap _{j\in J}{B}_j)=\bigcap _{j\in J}{f}^{-1}(B_j)$ (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 $S\subseteq X$

• $f(S)\cap f(T)=\mathrm{\varnothing }$ for all $S,T\subseteq X$ such that $S\cap T=\mathrm{\varnothing }$

• $f(S\setminus T)=f(S)\setminus f(T)$ for all $S,T\subseteq X$