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\rightline{Math 223\qquad Gottlieb}
\centerline{Functions Done Right}
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\noindent
A {\it map} $f\colon X\to Y$ is a rule $f$ which assigns to every object $x$ in the {\it source} $X$ an object $f(x)$ in the {\it target} $Y$.

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Definition 2:\ \ Two sets $X$ and $Y$ are {\it equal}, denoted $X=Y$, $\Leftrightarrow$ every object of $X$ is an object in $Y$ and every object in $Y$ is an object in $X$.

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Definition 3:\ \ Two maps are equal\ $\Leftrightarrow$ \ Their sources are equal, their targets are equal, and their rules do the same thing.

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In more symbolic notation this definition reads as follows:\ \ Let $f\colon X\to Y$ and $g\colon X'\to Y'$ be two maps .
Then $f=g\Leftrightarrow X=X'$, and $ Y=Y'$ and $f(a)=g(a)$ for all $a$ in the 
source.

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Definition 4.
Suppose we have two maps, $f\colon X\to Y$ and $g\colon Y\to Z$ (note the target of $f$ is the source of $g$).
Then the map $h\colon X\to Z$ is the {\it composition} of $f$ and $g\Leftrightarrow$ $h(x)=g(f(x))$ for every $x$ in the source $X$.

We say that $f(x)$ is the {\it image} of $x$, and that $x$ is a {\it preimage}
of $f(x)$. 
The set of all the preimages of $y$ we will call the
{\it fiber} of $y$.

The subset of the target $Y$ which consists of elements $y$ assigned to some
$x$ is called the {\it image} of $f$. Thus we can say that the {\it image} of $f$ is the subset of objects of the target with non--empty fibers.

If $A$ is a subset of the source $X$, then the {\it image} of $A$ in the target $Y$
consists of the set of images of all the objects in $A$. The image of $A$ is
denoted by $f(A)$. Similarly, the {\it preimage of a subset} $B$ of the target $Y$
is the set of all objects in the source $X$ whose images are objects in $B$.
The preimage of $B$ is denoted by $f^{-1}(B)$.

If the {\it image} of $f$ is the entire {\it target} of $f$, then we say that
$f$ is {\it onto}. Thus $f$ is onto if and only if no fiber is empty. 
If the fibers of $f$ have at most one object in it, then we
say $f$ is {\it one--to--one}. If every fiber of $f$ has exactly one object in
it, then we say that $f$ is {\it one--to--one onto} or equivalently that
$f$ is {\it bijective}.

\bye
