Let $X,Y$ be sets and $f:X\to Y$ be a function. For any $A\subseteq X$ , define $$f(A):=\lbrace f(x)\in Y \mid x\in A\rbrace$$ and any $B\subseteq Y$ , define $$f^{-1}(B):=\lbrace x\in X\mid f(x)\in B\rbrace.$$ 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 $A_1=\lbrace 1\rbrace$ , $A_2=\lbrace -1\rbrace$ . Equality occurs iff $f$ is one-to-one:

   Suppose $f(x)=f(y)=z$ . 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 $A_1$ 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 $A_2=\lbrace 2\rbrace$ .
5. $A\subseteq f^{-1}f(A)$ . Again, one finds that equality fails for the real function $f(x)=x^2$ by selecting $A=\lbrace 1\rbrace$ . 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 $\lbrace x,y\rbrace=f^{-1}f(\lbrace x,y\rbrace)=f^{-1}(\lbrace z\rbrace)$ . On the other hand $\lbrace x\rbrace =f^{-1}f(\lbrace x\rbrace)=f^{-1}(\lbrace z\rbrace)$ . So $\lbrace x,y\rbrace =\lbrace x\rbrace$ , $x=y$ .

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 and , 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 , $A\subseteq f^{-1}f(A)$ , so that $f(A)\subseteq ff^{-1}f(A)$ (by 1). Set $B=f(A)$ . Then by , $ff^{-1}f(A)=ff^{-1}(B)\subseteq B=f(A)$ .

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 $[f]$ 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 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.