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 one-to-one:
Suppose $f(x)=f(y)=z$ . Pick $A_{1}=\\{x\\}$ and $A_{2}=\\{y\\}$ . Then $f(A_{1}\\cap A_{2})=f(A_{1})\\cap f(A_{2})=\\{z\\}\eq\\varnothing$ . This means that $A_{1}\\cap A_{2}\eq\\varnothing$ . Since both $A_{1}$ and $A_{2}$ are singletons, $A_{1}=A_{2}$ , or $x=y$ .
Conversely, let’s show that $f$ 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 $x=f(a_{1})=f(a_{2})$ . As $f$ is one-to-one, $a_{1}=a_{2}$ 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.
$f^{-1}(Y-B)=X-f^{-1}(B)$ . As a result, $f^{-1}(B_{1}-B_{2})=f^{-1}(B_{1})-f^{-1}(B_{2})$ .
10.
$ff^{-1}(B)\\subseteq B$ . 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 $y=f(x)\\in ff^{-1}(B)$ . Now suppose the convserse, then pick $B=Y$ , and we have $Y=ff^{-1}(Y)=f(X)$ .
11.
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 $B=f(A)$ . Then by 10, $ff^{-1}f(A)=ff^{-1}(B)\\subseteq B=f(A)$ .

Remarks.

•
$f^{-1}f$ and $ff^{-1}$ 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 $ff^{-1}(B)=f(f^{-1}(B))$ .
•
From the definition above, we see that a function $f:X\\to Y$ induces two functions $[f]$ and $[f^{-1}]$ defined by
$[f]:2^{X}\\to 2^{Y}\\mbox{ such that }[f](A):=f(A)\\mbox{ and}$
$[f^{-1}]:2^{Y}\\to 2^{X}\\mbox{ such that }[f^{-1}](B):=f^{-1}(B).$
The last property 11 says that $[f]$ and $[f^{-1}]$ are quasi-inverses of each other.
•
$f$ is a bijection iff $[f]$ and $[f^{-1}]$ are inverses of one another.