PlanetMath: properties of a function

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (195)
Orphanage
Unclass'd
Unproven (498)
Corrections (25)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

properties of a function

(Definition)

Let be sets and $f:X\to Y$ be a function. For any $A\subseteq X$ , define

$\displaystyle f(A):=\lbrace f(x)\in Y \mid x\in A\rbrace$

and any $B\subseteq Y$ , define

$\displaystyle f^{-1}(B):=\lbrace x\in X\mid f(x)\in B\rbrace.$

is a subset of

and $f^{-1}(B)$ is a subset of

Let be arbitrary subsets of and be arbitrary subsets of , where belongs to the index set and to the index set . We have the following properties:

If $A_1\subset A_2$ , then $f(A_1)\subseteq f(A_2)$ . In particular, $f(A)\subseteq f(X)$ .
$f(A_1\cup A_2)=f(A_1)\cup f(A_2)$ . More generally, $f(\bigcup_i A_i)=\bigcup_i f(A_i)$ .
$f(A_1\cap A_2)\subseteq f(A_1)\cap f(A_2)$ . The equality fails in the example where is a real function defined by and $A_1=\lbrace 1\rbrace$ , $A_2=\lbrace -1\rbrace$ . Equality occurs iff is one-to-one:
Suppose . Pick $A_1=\lbrace x\rbrace$ and $A_2= \lbrace y\rbrace$ . Then $f(A_1\cap A_2) = f(A_1)\cap f(A_2)= \lbrace z\rbrace\ne\varnothing$ . This means that $A_1\cap A_2\ne \varnothing$ . Since both and are singletons, , or .

Conversely, let's show that is one-to-one then $f(A_1\cap A_2) = f(A_1)\cap f(A_2)$ . To do this, we only need to show the right hand side is included in the left, and this follows since if $x \in f(A_1)\cap f(A_2)$ then for some $a_1 \in A_1$ and $a_2 \in A_2$ we have . As is one-to-one, and so lies in $A_1 \cap A_2$ and is in $f(A_1 \cap A_2)$ .

More generally, $f(\bigcap_i A_i)\subseteq \bigcap_i f(A_i)$ .
$f(A_1)-f(A_2)\subseteq f(A_1-A_2)$ : If $y\in f(A_1)-f(A_2)$ , then for some $x\in A_1$ . If $x\in A_2$ , then $y=f(x)\in f(A_2)$ as well, a contradiction. So $x\in A_1-A_2$ , and $y=f(x)\in f(A_1-A_2)$ . The inequality is strict in the case when $f:\mathbb{Z}\to \mathbb{Z}$ given by , and $A_1=\mathbb{Z}$ and $A_2=\lbrace 2\rbrace$ .
$A\subseteq f^{-1}f(A)$ . Again, one finds that equality fails for the real function by selecting $A=\lbrace 1\rbrace$ . Equality again holds iff is injective:
Suppose $x \in f^{-1}f(A)$ . By definition this means that for some $x \in A$ , and since is injective we have $x = a \in A$ . It follows that $f^{-1}f(A) \subseteq A$ . Convserly, if , then $\lbrace x,y\rbrace=f^{-1}f(\lbrace x,y\rbrace)=f^{-1}(\lbrace z\rbrace)$ . On the other hand $\lbrace x\rbrace =f^{-1}f(\lbrace x\rbrace)=f^{-1}(\lbrace z\rbrace)$ . So $\lbrace x,y\rbrace =\lbrace x\rbrace$ , .
If $B_1\subseteq B_2$ , then $f^{-1}(B_1)\subseteq f^{-1}(B_2)$ . In particular, $f^{-1}(B)\subseteq f^{-1}(Y)$ .
$f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$ . More generally, $f^{-1}(\bigcup_j B_j)=\bigcup_j f^{-1}(B_j)$ .
$f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)$ . More generally, $f^{-1}(\bigcap_j B_j)=\bigcap_j f^{-1}(B_j)$ .
$f^{-1}(Y-B)=X-f^{-1}(B)$ . As a result, $f^{-1}(B_1-B_2)=f^{-1}(B_1)-f^{-1}(B_2)$ .
$ff^{-1}(B) \subseteq B$ . Yet again, one finds that equality fails for the real function by selecting . Equality holds iff is surjective:
Suppose is onto. Pick any $y\in B\subset Y$ . Then for some $x\in X$ . In other words, $x\in f^{-1}(B)$ and hence $y=f(x)\in ff^{-1}(B)$ . Now suppose the convserse, then pick , and we have $Y=ff^{-1}(Y)=f(X)$ .
Combining 10 and 5, we have that $ff^{-1}f(A)=f(A)$ and $f^{-1}ff^{-1}(B)=f^{-1}(B)$ . Let's show the first equality:
From 5, $A\subseteq f^{-1}f(A)$ , so that $f(A)\subseteq ff^{-1}f(A)$ (by 1). Set . Then by 10, $ff^{-1}f(A)=ff^{-1}(B)\subseteq B=f(A)$ .

Remarks.

$f^{-1}f$ and $ff^{-1}$ mean the compositions of the function and its inverse as defined at the beginning of the entry, so that $f^{-1}f(A)=f^{-1}(f(A))$ and $ff^{-1}(B)=f(f^{-1}(B))$ .
From the definition above, we see that a function $f:X\to Y$ induces two functions and $[f^{-1}]$ defined by
$\displaystyle [f]:2^X\to 2^Y$ such that $\displaystyle [f](A):=f(A)$ and

$\displaystyle [f^{-1}]:2^Y\to 2^X$ such that $\displaystyle [f^{-1}](B):=f^{-1}(B).$
The last property 11 says that and $[f^{-1}]$ are quasi-inverses of each other.
is a bijection iff and $[f^{-1}]$ are inverses of one another.