First, I wan't to make clear that we use the same terms:
$X, Y$ are sets with $X \neq \emptyset$ and $Y \neq \emptyset$.
$f:X \rightarrow Y$ is a funtion.
$f^{-1}(A)$ is the preimage of $A \subseteq Y$ (and NOT the inverse function).
$f$ is a function from domain $X$ to codomain $Y$. The smaller oval inside $Y$ is the image of $f$.
Source: commons.wikimedia.org
Question:
What is the preimage of the codomain of a function if the function is not surjective?
Example:
$f: \mathbb{R} \rightarrow \mathbb{R}$
$f(x) := x^2$
a) $f^{-1}(\{-1,1,2\}) = \{1, \sqrt{2}, -1, \sqrt{2}\}$ and especially $f^{-1}(\{-1\}) = \emptyset$ or
b) $f^{-1}(\{-1,1,2\})$ is undefined, as $-1$ has no preimage?
Which one is correct? Does it depend on the author?
Best Answer
Definitely (a). The definition of preimage is $f^{-1}(A) = \{x \in X : f(x) \in A\}$ so it's defined for all subsets $A \subset Y$, regardless of how $A$ relates to the image of $f$.