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\centerline{Mathematics 243, section 1 -- Algebraic Structures}
\centerline{Discussion 2 -- Composition Inverses}
\centerline{September 19, 2003}
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\noindent
{\it Background}
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Last time we discussed the binary operation of composition
$\circ$ on the set $A = F({\bf R}) = \{f : f : {\bf R} \to {\bf R}\}$.
We also introduced the following general terminology for
binary operations which have identity elements:
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\item{$\bullet$} We say $b \in A$ is a {\it left inverse of } $a \in A$
if $b*a = e$ (the identity element).
\item{$\bullet$} We say $b \in A$ is a {\it right inverse of } $a \in A$
if $a*b = e$.
\item{$\bullet$} We say $b \in A$ is an {\it inverse of } $a \in A$
if $a*b = b*a = e$.
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\noindent
Today, we want to tie these notions together with some of the 
function concepts (ranges, injectivity, surjectivity, etc.) 
that we learned last week.

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\noindent
{\it Note:}  The facts we will be considering today are covered in 
the text.  You may refer to the discussions in section 1.3 after
class if you need to, but at least for now, try to ``figure this 
out on your own'' -- it will be very good practice!
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\noindent
{\it Discussion Questions}
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\item{A)} Of the following four functions in $F({\bf R})$, 
one has neither a left nor a right (composition) inverse in $F({\bf R})$, 
one has a left inverse but no right inverse,  one has a right inverse
but no left inverse, and one has a left and a right inverse (which 
are equal).  Explain which is which and give formulas for the 
left or right inverse functions that do exist.  {\it Note}:
Your left and right inverse functions should be in $F({\bf R})$ -- 
that is the {\it domains should be the whole set of real numbers} in all cases.
\itemitem{1)}  $f(x) = \arctan(x)$
\itemitem{2)}  $g(x) = x^4$
\itemitem{3)}  $h(x) = {\root 5\of {x + 7}}$
\itemitem{4)}  $i(x) = \cases{\ln(|x|) & if $x \ne 0$\cr
                                   0   & if $x= 0$\cr}$.

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\item{B)}  If a mapping $f: A \to A$ has a right inverse $g : A \to A$
for composition, 
then $f(g(a)) = a$ for all $a \in A$.  In particular, note
that this says $f$ is {\it surjective}.  Say why this is true, 
and show this observation agrees with your answers in question A
(with $A  = {\bf R}$ for those functions).

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\item{C)}  Now, suppose we know that $f : A \to A$ is surjective.  We can show
that in fact $f$ must have a right composition inverse as follows.
Since $f$ is surjective, for all $b \in A$, there exist
some $a\in A$ such that $f(a) = b$.  There may in fact be
more than one such $a$, but just choose any one that ``works'' and
let $g(b) = $ that $a$.  Show that the mapping defined this way
is a right composition inverse for $f$.

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\item{D)}  Saying that a mapping $f : A \to A$ has a left inverse 
for composition says that there is some mapping $g : A \to A$ such that 
$g \circ f = I_A$
(the identity function on $A$).  If you have an equation 
$f(a) = f(b)$ for some $a, b \in A$
and you apply the left inverse $g$ to both sides, what 
happens?  What does this imply about $f$?  Does this agree with
what you saw in question A?

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\item{E)}  Your answer to question B should be equivalent to 
an implication $f$ has a left composition inverse $\Rightarrow$ $f$ has some
other ``standard'' property we have discussed.  The converse statement:
$f$ has the ``standard property'' $\Rightarrow$ $f$ has 
a left composition inverse is also true.  Say how to construct
a left inverse if $f$ has the ``standard property.''  (Hint: There
is no choice for where the left inverse maps 
$b \in {\rm range}(f) \subseteq A$.
The choice of what the inverse does on the complement of the 
range is actually arbitrary, though.  Do you see why?  Think about 
function $i(x)$ from part A.)

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\noindent
{\it Assignment}
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Group write-ups due in class Wednesday, September 24.
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