# Definition Consider a [[Map|map]] $f:X\rightarrow Y$. The *domain* is $M$ and the *co-domain* is $N$. Consider $S \subset X$, the elements in $Y$ mapped from points of $S$ form a set $T \subset Y$ called the *image* or the *range* of $S$ under $f$, denoted by $f(S)$. Conversely, the set $S$ is called the *inverse image* of $T$, denoted by $f^{-1}(T)$. If the [[map]] is *many-to-one*, then the inverse image of a single point of $Y$ is not a single point of $X$, so there is no *[[Map]]* $f^{-1}$ from $Y$ to $X$ (recall that a [[map]] must map each element in its domain to a unique element in its range). In this case, the symbol $f^{-1}(T)$ must be read as a single symbol, $f^{-1}(T)$ is simply a set with that name, not the image of $T$ under a [[map]] $f^{-1}$ In other words, consider a subset of $X$ whose elements are mapped to $y \in Y$ under $f$. This subset is called the *inverse image* or *preimage* of $y$, denoted by $f^{-1}(y) = \{x \in X | f(x) = y\} \subset X$. On the other hand, the image of the map is $\text{im}(f) = f(X) = \{y \in Y | y = f(x) \text{ for some } x \in X\} \subset Y$ The domain and co-domain are part of the map's definition. # Example 1 ![[Codomain2.svg]] Here, the red blob is the *domain*, the blue blob is the *co-domain*, and the yellow blob is the *image* or the *range*. Image from Wikipedia. # Example 2 Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = \sin(x)$. Then, the domain is $\mathbb{R}$, the co-domain is $\mathbb{R}$, and the range or image is $[-1,1]$. The inverse image of $0$ is $f^{-1}(0) = \{n\pi|n \in \mathbb{Z}\}$