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\begin{flushright}
Introduction to Abstract Mathematics \\
MATH 310\\
Homework 10\\
Due Monday December 7
\end{flushright}

\noindent
Please show all your work and justify your answers.

\begin{exer}
	Let $A = \{5,6\}$, $B = \{5,7,8\}$, and $S = \{ n\in \mathbb{Z}\mid n\geq 3 \text{ is odd}\}$. A relation $R$ from $A\times B$ to  $S$ is defined as $(a,b) \, R \, s$ if $s\mid (a+b)$. Is $R$ a function from $A\times B$ to $S$?
\end{exer}

\begin{exer}
	For a function $f:A\to B$ and subsets $C$ and $D$ of $A$ and $E$ and $F$ of $B$, prove the following.
	\begin{enumerate}[(a)]
	\item $f(C\cup D) = f(C) \cup f(D)$
	\item $f(C\cap D) \subseteq f(C)\cap f(D)$
	\item $f(C) \setminus f(D) \subseteq f(C \setminus D)$
	\item $f^{-1}(E\cup F) = f^{-1}(E) \cup f^{-1}(F)$
	\item $f^{-1}(E\cap F) = f^{-1}(E) \cap f^{-1}(F)$
	\item $f^{-1}(E\setminus F) = f^{-1}(E) \setminus f^{-1}(F)$.
	\end{enumerate}
\end{exer}

\begin{exer}
	Give an example of two finite sets $A$ and $B$ and two functions $f:A\to B$ and $g:B\to A$ such that $f$ is one-to-one but not onto and $g$ is onto but not one-to-one.
\end{exer}

\begin{exer}
	Let $f$ be a function with $\operatorname{dom}(f) = A$ and let $C$ and $D$ be subsets of $A$. Prove that if $f$ is one-to-one, then $f(C\cap D) = f(C)\cap f(D)$.
\end{exer}

\begin{exer}
	Let $A$ and $B$ be nonempty sets. Prove that if $f:A\to B$ is a function, then $f\circ i_A = f$ and $i_B\circ f = f$.
\end{exer}

\begin{exer}
	Let $A$ be a nonempty set and let $f:A\to A$ be a function. Prove that if $f\circ f = i_A$, then $f$ is bijective.
\end{exer}

\begin{exer}
	Let $A,B$ and $C$ be nonempty sets and let $f,g$ and $h$ be functions such that $f:A\to B$, $g:B\to C$ and $h:B\to C$. For each of the following, prove or disprove:
	\begin{enumerate}[(a)]
	\item If $g\circ f = h\circ f$, then $g=h$.
	\item If $f$ is one-to-one and $g\circ f = h\circ f$, then $g=h$.
	\end{enumerate}
\end{exer}

\begin{exer}
	Suppose, for a function $f:A\to B$, there exists a function $g:B\to A$ such that $f\circ g = i_B$. Prove that if $g$ is surjective, then $g\circ f = i_A$.
\end{exer}


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