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\begin{center}
{\large MATH 108 Fall 2019 \ - \ {\bf Problem Set 7}}
% Put your name and section here.
{\large due November 15}
\end{center}
\begin{enumerate}
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\item For each function $f$, determine if it is surjective. If yes, find a {\em right-inverse} of $f$, which is a function $g$ such that $f \circ g$ is the identity.
\begin{enumerate}
\item $f:\RR \to \RR^2$ defined by $f(x) = (x,x)$.
\item $f:\RR^2 \to \RR$ defined by $f(x,y) = x+y$.
\item $f:\ZZ \to \ZZ/4\ZZ$ defined by $f(x) = \overline{x}$.
\item $f:\RR \to \RR$ defined by $f(x) = e^x$.
\item $f:\ZZ \to \{0\}$ defined by $f(x) = 0$.
\end{enumerate}
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\item Let $f:A \to B$ and $g:B \to C$.
\begin{enumerate}
\item Prove that if $g \circ f$ is surjective then $g$ is surjective.
\item Give an example of $f$ and $g$ where $g \circ f$ is surjective but $f$ is not surjective.
\end{enumerate}
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\item Prove that each function is a bijection. Give the inverse.
\begin{enumerate}
\item $f:\ZZ \to \ZZ$ defined by $f(x) = x+1$.
\item $f:(2,\infty) \to (-\infty,-1)$ defined by $f(x) = \dfrac{-x}{x-2}$.
\item $f:\ZZ/8\ZZ \to \ZZ/8\ZZ$ defined by $f(\overline{x}) = \overline{5x-1}$.
\end{enumerate}
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\item For each pair of sets, find a bijection from the first to the second.
\begin{enumerate}
\item $\ZZ_{>0}$ and $\ZZ_{\geq 0}$.
\item $\RR^2$ and $\CC$.
\item $\ZZ$ and $\ZZ_{>0}$.
\item $\{x \in \RR \mid -1 < x < 1\}$ and $\RR$.
\end{enumerate}
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\item For postive integers $n$ and $m$, let $[n] = \{1,2,\ldots,n\}$ and $[m] = \{1,2,\ldots,m\}$.
\begin{enumerate}
\item Let $A$ be the set of all functions from $[n]$ to $[m]$. Compute $|A|$ in terms of $n$ and $m$.
\item Let $B$ be the set of all bijective functions from $[n]$ to $[m]$. Compute $|B|$ in terms of $n$ and $m$.
\item Let $C$ be the set of all injective functions from $[n]$ to $[m]$. Compute $|C|$ in terms of $n$ and $m$.
\end{enumerate}
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\item Let $f_1,f_2:A\to B$ and $g:B\to C$ and $h_1,h_2:C \to D$.
\begin{enumerate}
\item Prove that if $g\circ f_1 = g \circ f_2$ and $g$ is injective, then $f_1 = f_2$.
\item Prove that if $h_1 \circ g = h_2 \circ g$ and $g$ is surjective, then $h_1 = h_2$.
\end{enumerate}
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\end{enumerate}
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