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 nonempty index sets^{}.
Then, it is easy to prove, directly from definitions, that the following hold:

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

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$f(\bigcap _{i\in I}{A}_{i})\subseteq \bigcap _{i\in I}f({A}_{i})$ (i.e., the image of an intersection^{} is contained in the intersection of the images)

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$A\subseteq {f}^{1}(f(A))$ for any $A\subseteq X$ (where ${f}^{1}(f(A))$ is the inverse image^{} of $f(A)$)

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$f({f}^{1}(B))\subseteq B$ for any $B\subseteq Y$

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${f}^{1}(Y\setminus B)=X\setminus {f}^{1}(B)$ for any $B\subseteq Y$

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

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${f}^{1}(\bigcap _{j\in J}{B}_{j})=\bigcap _{j\in J}{f}^{1}({B}_{j})$ (the inverse image of an intersection is the intersection of the inverse images)

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$f({f}^{1}(B))=B$ 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):

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$f$ is injective

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

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${f}^{1}(f(S))=S$ for all $S\subseteq X$

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

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$f(S\setminus T)=f(S)\setminus f(T)$ for all $S,T\subseteq X$
Title  properties of functions^{} 

Canonical name  PropertiesOfFunctions 
Date of creation  20130322 14:59:54 
Last modified on  20130322 14:59:54 
Owner  yark (2760) 
Last modified by  yark (2760) 
Numerical id  20 
Author  yark (2760) 
Entry type  Result 
Classification  msc 03E20 
Related topic  PropertiesOfAFunction 