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\centerline{Mathematics 243, section 1 -- Algebraic Structures}
\centerline{Discussion 1: More on Mappings}
\centerline{September 12, 2003}
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{\it Background}
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Last time, we introduced the following terminology for mappings (functions)
$f : A \to B$:
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\item{1)}  The {\it domain\/} of $f$ is the set $A$.
\item{2)}  The {\it codomain\/} of $f$ is the set $B$.
\item{3)}  The {\it range\/} of $f$ is the set $f(A) = \{b \in B | b = f(a) {\rm\ 
for\ some\ } a \in A\}$.
\item{4)}  The function $f$ is {\it surjective, (or onto)\/} if
$f(A) = B$ (or equivalently, if for every $b \in B$, there exists
some $a \in A$ such that $f(a) = b$).
\item{5)}  The function $f$ is {\it injective, or (one-to-one)\/} if
$f(a) = f(a')$ implies $a = a'$ (or stated yet another way, if 
distinct elements $a \ne a'$ of $A$ map to distinct elements 
$f(a) \ne f(a')$ of $B$).
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Today we will practice using these ideas with some specific examples of
functions.  Before we begin, one comment about the definitions of the 
terms ``surjective'' and ``injective'' is probably in order.  
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{\bf Important Comment:} It is conventional in making definitions of 
mathematical concepts or categories of objects to 
state the definition in the form: ``$x$ is a {\it foo\/} if it has
the property {\it bar\/}".  This is one case where the intended meaning 
is {\it not given by a straight translation\/} into symbolic logical
form.  The symbolic form would be
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\centerline{$x$ has property {\it bar} $\Rightarrow$ $x$ is a {\it foo}}
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But that is {\bf not} how you should read the definition.
The definition really means:
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\centerline{$x$ has property {\it bar} $\Leftrightarrow$ $x$ is a {\it foo}}
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In other words, for example, the definition of the surjectivity property 
for functions really means {\it both}:  ``if $f(A) = B$, then $f$
is surjective'' {\it and} ``if $f$ is surjective, then $f(A) = B$''.
Why do you suppose people do not include the other implication explicitly?
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{\it Discussion Questions}
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\item{A)}  For each of the following functions from ${\bf R}$ to ${\bf R}$, 
the domain is the largest set of $x \in {\bf R}$ for which the formula makes 
sense.  Say what the domain is, what the range is, and determine whether $f$
is injective, surjective, both, or neither.
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\itemitem{1)}  $f$ defined by $f(x) = x^2 + 5x + 6$
\itemitem{2)}  $g$ defined by $g(x) = \sqrt{1 - x^2}$
\itemitem{3)}  $h$ defined by $h(x) = \arctan(x)$ (principal value of 
inverse tangent)
\itemitem{4)}  $i$ defined by $i(x) = {1\over x^2 - 3x}$
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\item{B)}  The {\it composition} of two functions $f : A \to B$ 
and $g : B \to C$ is the function $g \circ f : A \to C$ defined by 
$(g \circ f)(a) = g(f(a))$ for all $a \in A$.
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\itemitem{1)}  What are formulas that define the compositions $f\circ h$ and
$h \circ f$ using the functions from question A?
\itemitem{2)}  In general, now, if $f$ and $g$ are both injective, 
prove that $g\circ f$ is injective.
\itemitem{3)}  If $f$ and $g$ are both surjective, does it follow that
$g\circ f$ is surjective?  Prove, or give a counterexample.
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\item{C)}  Let $A = \{1,2,3\}$ and $B = \{a,b,c\}$.
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\itemitem{1)} How many different functions $f : A \to B$ are there?
(Explain -- what does it mean to say two functions are equal, or different?
How are you counting the functions?  How could you construct
all of them?)
\itemitem{2)}  How many of the functions from part 1 are {\it injective\/}?
List the ones that are.
\itemitem{3)}  How many of the functions from part 1 are {\it surjective\/}?
Again, list the ones that are.
\itemitem{4)}  Your answers to parts 2 and 3 of this question should
be {\it the same}.  Explain why that is true in intuitive terms (that is,
it is not necessary to try to write out a formal proof).
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{\it Assignment}
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Group write-ups due in class Wednesday, September 17.
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