%EXERCISE NINE
          
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\begin{center}
	\textbf{Exercise Nine: Arrows and Functions}
\end{center}  

\begin{dfn}
A function $f\colon A\to B$ is \emph{bijective} or a \emph{one-to-one correspondence} 
if and only if $f$ is injective and surjective.
\end{dfn}

\begin{notation}
If $f$ is a bijection, we write 
$f\colon A\xrightarrow[\text{onto}]{\text{1-1}} B$.
\end{notation}

\begin{clm}
The function $f\colon\mathbb{R}\to\mathbb{R^+}$ given by $f(x)=x^4+1$ is not injective.
\end{clm}

\begin{proof}
We must show that there exist $a_1, a_2\in\mathbb{R}$ such that $f(a_1)=f(a_2)$ 
but $a_1\neq a_2$.

Choose $a_1=1$ and $a_2=-1$. 

Then $f(a_1)=2$ and $f(a_2)=2$ but $a_1\neq a_2$, so $f$ is not injective.\\
\end{proof}

\begin{clm}
The function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x)=3x+1$ is surjective.
\end{clm}

\begin{proof}
We must show that for every $b\in\mathbb{R}$ [the codomain] there exists
$a\in\mathbb{R}$ [the domain] such that $f(a)=b$.

Pick any $b\in\mathbb{R}$.

Let $a=(b-1)/3$. 

Since $b\in\mathbb{R}$ and because the reals are closed under subtraction and non-zero
division, we know that $(b-1)/3\in\mathbb{R}$, \emph{i.e.}, $a$ is in the domain of $f$.

Furthermore, 
\begin{align*}
f(a) &= f\left(\frac{b-1}{3}\right) \\
% Recall \left and \right from Lesson Six
&= 3\cdot\frac{b-1}{3}+1\\
&= b-1+1\\
&= b 
\end{align*}
as desired.
\end{proof}

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