properties of a function
Let $X,Y$ be sets and $f:X\to Y$ be a function. For any $A\subseteq X$, define
$$f(A):=\{f(x)\in Y\mid x\in A\}$$ 
and any $B\subseteq Y$, define
$${f}^{1}(B):=\{x\in X\mid f(x)\in B\}.$$ 
So $f(A)$ is a subset of $Y$ and ${f}^{1}(B)$ is a subset of $X$.
Let $A,{A}_{1},{A}_{2},{A}_{i}$ be arbitrary subsets of $X$ and $B,{B}_{1},{B}_{2},{B}_{j}$ be arbitrary subsets of $Y$, where $i$ belongs to the index set^{} $I$ and $j$ to the index set $J$. We have the following properties:

1.
If ${A}_{1}\subset {A}_{2}$, then $f({A}_{1})\subseteq f({A}_{2})$. In particular, $f(A)\subseteq f(X)$.

2.
$f({A}_{1}\cup {A}_{2})=f({A}_{1})\cup f({A}_{2})$. More generally, $f({\bigcup}_{i}{A}_{i})={\bigcup}_{i}f({A}_{i})$.

3.
$f({A}_{1}\cap {A}_{2})\subseteq f({A}_{1})\cap f({A}_{2})$. The equality fails in the example where $f$ is a real function defined by $f(x)={x}^{2}$ and ${A}_{1}=\{1\}$, ${A}_{2}=\{1\}$. Equality occurs iff $f$ is onetoone:
Suppose $f(x)=f(y)=z$. Pick ${A}_{1}=\{x\}$ and ${A}_{2}=\{y\}$. Then $f({A}_{1}\cap {A}_{2})=f({A}_{1})\cap f({A}_{2})=\{z\}\ne \mathrm{\varnothing}$. This means that ${A}_{1}\cap {A}_{2}\ne \mathrm{\varnothing}$. Since both ${A}_{1}$ and ${A}_{2}$ are singletons, ${A}_{1}={A}_{2}$, or $x=y$.
Conversely, let’s show that $f$ is onetoone 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 $x=f({a}_{1})=f({a}_{2})$. As $f$ is onetoone, ${a}_{1}={a}_{2}$ and so ${a}_{1}$ lies in ${A}_{1}\cap {A}_{2}$ and $x$ is in $f({A}_{1}\cap {A}_{2})$.
More generally, $f({\bigcap}_{i}{A}_{i})\subseteq {\bigcap}_{i}f({A}_{i})$.

4.
$f({A}_{1})f({A}_{2})\subseteq f({A}_{1}{A}_{2})$: If $y\in f({A}_{1})f({A}_{2})$, then $y=f(x)$ 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 $f(x)=1$, and ${A}_{1}=\mathbb{Z}$ and ${A}_{2}=\{2\}$.

5.
$A\subseteq {f}^{1}f(A)$. Again, one finds that equality fails for the real function $f(x)={x}^{2}$ by selecting $A=\{1\}$. Equality again holds iff $f$ is injective:
Suppose $x\in {f}^{1}f(A)$. By definition this means that $f(x)=f(a)$ for some $x\in A$, and since $f$ is injective we have $x=a\in A$. It follows that ${f}^{1}f(A)\subseteq A$. Convserly, if $f(x)=f(y)=z$, then $\{x,y\}={f}^{1}f(\{x,y\})={f}^{1}(\{z\})$. On the other hand $\{x\}={f}^{1}f(\{x\})={f}^{1}(\{z\})$. So $\{x,y\}=\{x\}$, $x=y$.

6.
If ${B}_{1}\subseteq {B}_{2}$, then ${f}^{1}({B}_{1})\subseteq {f}^{1}({B}_{2})$. In particular, ${f}^{1}(B)\subseteq {f}^{1}(Y)$.

7.
${f}^{1}({B}_{1}\cup {B}_{2})={f}^{1}({B}_{1})\cup {f}^{1}({B}_{2})$. More generally, ${f}^{1}({\bigcup}_{j}{B}_{j})={\bigcup}_{j}{f}^{1}({B}_{j})$.

8.
${f}^{1}({B}_{1}\cap {B}_{2})={f}^{1}({B}_{1})\cap {f}^{1}({B}_{2})$. More generally, ${f}^{1}({\bigcap}_{j}{B}_{j})={\bigcap}_{j}{f}^{1}({B}_{j})$.

9.
${f}^{1}(YB)=X{f}^{1}(B)$. As a result, ${f}^{1}({B}_{1}{B}_{2})={f}^{1}({B}_{1}){f}^{1}({B}_{2})$.

10.
$f{f}^{1}(B)\subseteq B$. Yet again, one finds that equality fails for the real function $f(x)={x}^{2}$ by selecting $B=[1,1]$. Equality holds iff $f$ is surjective^{}:
Suppose $f$ is onto. Pick any $y\in B\subset Y$. Then $y=f(x)$ for some $x\in X$. In other words, $x\in {f}^{1}(B)$ and hence $y=f(x)\in f{f}^{1}(B)$. Now suppose the convserse, then pick $B=Y$, and we have $Y=f{f}^{1}(Y)=f(X)$.
 11.
Remarks.

•
${f}^{1}f$ and $f{f}^{1}$ 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 $f{f}^{1}(B)=f({f}^{1}(B))$.

•
From the definition above, we see that a function $f:X\to Y$ induces two functions $[f]$ and $[{f}^{1}]$ defined by
$$[f]:{2}^{X}\to {2}^{Y}\text{such that}[f](A):=f(A)\text{and}$$ $$[{f}^{1}]:{2}^{Y}\to {2}^{X}\text{such that}[{f}^{1}](B):={f}^{1}(B).$$ The last property 11 says that $[f]$ and $[{f}^{1}]$ are quasiinverses of each other.

•
$f$ is a bijection iff $[f]$ and $[{f}^{1}]$ are inverses of one another.
Title  properties of a function 

Canonical name  PropertiesOfAFunction 
Date of creation  20130322 16:21:38 
Last modified on  20130322 16:21:38 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  24 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03E20 
Related topic  PropertiesOfFunctions 