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\date{01/20/2022}
\title{%
            Set Theory: Functions\\
            \large MATH--190--01 : Discrete Mathematics for Computing}
\author{Blizzard MacDougall}
\begin{document}
\maketitle
\pagenumbering{arabic}
Functions work just like in math before.
Its denoted as:
\begin{equation}
    f:X\to Y
\end{equation}
This is an assignment of X to Y, such that every element of $X$ is assigned to
exactly one element $Y$. This is also written as $X\mapsto Y$, or $x\xrightarrow{f}y$.
If $X\mapsto Y$ is a function, then $f(x)$, is called \say{f of x} or \say{the image of x under f}.

X is the domain, Y is the \emph{codomain}. The range is $\{y\in Y | y=f(x)\ for\ some\ x\in X\}$.
Note that \verb|range|$(f)\subseteq Y$, but it is not necessarily equal.

For a function $f: X \mapsto Y$, given a $y\in Y$ there \emph{may} be a function such that
$f(x)=y$. If there is such an $x$, it is called the \emph{preimage} or \emph{inverse image} of $y$.
There may be multiple of these $x$s.
The inverse image of y is $\{x\in X|f(x)=y\}$

A function is one-to-one or \emph{injective} if there is only one $x$ mapped to every $y$.
Functions that are not injective collapse points together.

To prove injectivity, either pick $x_1$ and $x_2$ and prove that $[f(x_1)=f(x_2)]\to (x_1=x_2)$,
or pick two arbitrary elements $x_1\neq x_2$ and show that $f(x_1)\neq f(x_2)$

A function $f: X\mapsto Y$ is onto or \emph{surjective} if every $y$ is the image of some $x$.
That is, the range is the same as the codomain.

A function $f:X\mapsto Y$ is a bijection if $f$ is \emph{both} injective and surjective.
This is also called one-to-one correspondence.

$f^{-1}(x)=y \leftrightarrow f(y)=x$

If $f$ is a bijection, then $f^{-1}$ is a bijection.
\end{document}