Can we call domain as inverse image of a function? I was going through the definition of inverse image of a function http://www.northeastern.edu/suciu/U565/MATH4565-sp10-handout1.pdf, and I was wondering if inverse image of a function is the domain of the function itself. Please give me some examples   on it.
 A: Hint: Given a function $f$ and a set $U$, the inverse image of $U$ under $f$ (it is necessary to specify both the set and the function, here) will be the domain of $f$ if and only if $U$ contains the image of $f$ (that is, $U$ is a codomain of $f$).
For an example to see why the set is important, let $f(x)=|x|.$ The inverse image of $\{1\}$ under $f$ is $\{-1,1\},$ but the inverse image of $\Bbb R$ under $f$ is $\Bbb R.$
A: "Inverse image of a function" doesn't make much sense( unless it is taken as inverse image of the co-domain ); you need to specify some subset of co-domain and then we can talk about "inverse image of set $B\subset$ co-domain w.r.t. function $f$".
Consider a function $f:A\to B$. Let $C\subset A$ and $D\subset B$ such that $f(C)=D$ i.e. $D=$ the set of elements of $B$ to which elements of $C$ are mapped.
Then inverse image of $D$ is precisely the subset of $A$ which is mapped to $D$ which is none other than $C$ 
Here $f(C)=D$ and $f^{-1}(D)=C$ which might not be the whole $A$, in fact, $f^{-1}(D)=A\iff f(A)=D$    
A: Let $f$ be a function from $X$ to $Y$ commonly written $f:X \rightarrow Y$. If we define the inverse function $f^{-1}: Y \rightarrow X$ as $f^{-1}(y) = x$ if $f(x) = y$, then $f^{-1}(Y) \subset X = \text{Dom}(f)$. However, there need to be further conditions on $f$ for $f^{-1}(Y) = X$. That is, $f$ must be a bijection: Every element in $Y$ is an image of $f$ for some value $x \in X$, and $f(x_1) = f(x_2) \implies x_1 = x_2$.
For example, $f: \mathbb{R} \rightarrow \mathbb{R}^+ \cup \left\{ 0 \right\}$, $f(x) = x^2$ is a function that is not a bijection (for example, $f(-1) = f(1) = 1$). Likewise, $f^{-1} ( \mathbb{R}^+ \cup \left\{ 0 \right\} ) \subset \mathbb{R}^+ \cup \left\{ 0 \right\} \subset \mathbb{R}$. However, $f(x) = x^3$ on $\mathbb{R}$ is a bijection, and so the image of $R$ under $f^{-1}$ is $\mathbb{R} = \text{Dom}( f )$.
