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\handout{miniquiz7}{Mini-Quiz 7}{\tutdate}

\begin{enumerate}
\item Write your name:

\bigskip

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\item \textbf{Injections, Surjections, and Bijections}
%[[[Source:  first two parts are standard, second two are from
%Velleman, p. 231, Ex. 6]]]

Let $g$ be a function from $A$ to $B$, and $f$ a function from $B$ to
$C$.  The composition $f \circ g$ is defined by:
$f \circ g (x) = f(g(x))$.

%[[[Check that my notation is consistent with what's been used in the
%class, in Rosen, etc.]]]


Determine whether each of the following statements is true or false.
If it's true, prove it.  If it's false, give a counterexample.

\medskip
\begin{problemparts}

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\problempart
If $f$ and $g$ are both injective then $f \circ g$ is injective.


\solution{
\medskip
\textbf{Solution:}
Let $x$,$y$ be any two distint elements in $A$.  Since $g$ is one-to-one (surjective), $g(x)$ and $g(y)$ are distinct elements in $B$.  Since $f$ is one-to-one, $f(g(x))$ and $f(g(y))$ are distince elements of $C$.  Thus, $f \circ g$ is injective. 
}

\vspace{-\baselineskip}
\problempart
If $f$ and $g$ are both surjective then $f \circ g$ is surjective.

\solution{
\medskip
\textbf{Solution:}
Let $y$ be an arbitrary element in $C$.  Since $f$ is surjective, $y=f(b)$ for some element $b \in B$.  Since $g$ is also surjective, there must exist an $x \in A$ such that $b=g(x)$.  Therefore, for any element $y \in C$, there exists an element $x \in A$ such that $y=f(g(x))$.  Thus, $f \circ g$ is surjective.
}




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%\problempart
%If $g$ is injective and $g \circ f$ is surjective then $f$ is surjective.
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\end{problemparts}


\end{enumerate}

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