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:

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$f(\bigcup\limits_{i\in I}{A_{i}})=\bigcup\limits_{i\in I}{f(A_{i})}$ (i.e., the image of a union is the union of the images)
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$f(\bigcap\limits_{i\in I}{A_{i}})\subseteq\bigcap\limits_{i\in I}{f(A_{i})}$ (i.e., the image of an intersection is contained in the intersection of the images)
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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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$f^{-1}(\bigcup\limits_{j\in J}{B_{j}})=\bigcup\limits_{j\in J}{f^{-1}(B_{j})}$ (the inverse image of a union is the union of the inverse images)
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$f^{-1}(\bigcap\limits_{j\in J}{B_{j}})=\bigcap\limits_{j\in J}{f^{-1}(B_{j})}$ (the inverse image of an intersection is the intersection of the inverse images)
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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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$f(S)\cap f(T)=\varnothing$ for all $S,T\subseteq X$ such that $S\cap T=\varnothing$
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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	2013-03-22 14:59:54
Last modified on	2013-03-22 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