# 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. 1.

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

2. 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. 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 $f(A_{1}\cap A_{2})=f(A_{1})\cap f(A_{2})=\{z\}\neq\varnothing$. This means that $A_{1}\cap A_{2}\neq\varnothing$. 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. 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. 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. 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. 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. 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. 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. 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. 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.

Title properties of a function PropertiesOfAFunction 2013-03-22 16:21:38 2013-03-22 16:21:38 CWoo (3771) CWoo (3771) 24 CWoo (3771) Definition msc 03E20 PropertiesOfFunctions