Images and Preimages

Definition

$f (U) = {y \in B : \exists x \in U, f (x) = y} = {f (x) : x \in U}$
An image of $U$ is the $y$ from the Codomain where there is an $x$ in the image where the $f (x)$ is our $y$
An image of $U$ is the $y$ value from the graph where $f (x) = y$
Alternate way to think of it - as the range: $f (x) = {y \in B : y = f (x) for some x \in A}$

$f^{- 1} (V) = {x \in A : f (x) \in V}$
A preimage is where we have an $x$ from our Domain where $f (x)$ is part of our preimage.
A preimage is where the $x$ on the graph has a $y$ value that's part of the preimage.
$f^{- 1} (B) = A$

An Image is a subset of the Domain.
A Preimage is a subset of the Codomain.
- Example of images and pre images.
Example:
- $f : Z \to R, f (x) = x^{2}$
- Then the range of $f$ is $f (Z)$ .
- Basically take all inputs, and the computation on that input is the range, which is in the Codomain.
- This preimage is wrong here
  - $V = [0, 1]$ then $f^{- 1} (V) = [- 1, 1]$
  - Why? Because we include $[0, 1]$ totally. So this includes non-integers. So this preimage is incorrect unless the function changes.
  - When the function is $f : R \to R, f (x) = x^{2}$ , then the above preimage is correct.
Preimages
- A preimage respects unions, subsets, intersections, set complements, etc.
Images
- An image doesn't respect everything.
- Does respect unions
- Doesn't respect intersections
If $U \subseteq A$ then $U \subseteq f^{- 1} (f (U))$
If $V \subseteq B, f (f^{- 1} (v) \supseteq V)$