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\begin{document}

\noindent
{\bf Solutions to HW 4.}

\bigskip
  
\begin{enumerate}
\item[1.]
  Show that the kernel of a function with domain
  $A$ is an equivalence relation on $A$. 
\end{enumerate}
  
  \bigskip
  
\color{blue}
We must show that $\ker(f)$ is reflexive, symmetric, and transitive.
\begin{enumerate}
\item
  For any $a\in A$, $f(a)=f(a)$, so $(a,a)\in\ker(f)$.
  \medskip
  
\item
  For any $a, b\in A$, if $(a,b)\in\ker(f)$, then $f(a)=f(b)$.
  But then $f(b)=f(a)$, so $(b,a)\in\ker(f)$.
  \medskip
  
\item
  For any $a, b, c\in A$, if $(a,b), (b,c)\in\ker(f)$,
  then $f(a)=f(b)$ and $f(b)=f(c)$.
  But then $f(a)=f(c)$, so $(a,c)\in\ker(f)$.
\end{enumerate}
\color{black}

\vfill
  
\begin{enumerate}
\item[2.]
  Show that if $E$ is an equivalence relation on $A$,
  then $E$ is the kernel of some function with domain $A$.
  (Hint: You need to find a function with domain $A$ and kernel $E$.
  Let $P_E=\{[a]\in E\;|\;a\in A\}$ be the partition associated to
  $E$. Show that the natural map $\nu:A\to P_E:a\mapsto [a]$ has kernel $E$.)
\end{enumerate}
  

\bigskip

\color{blue}
Following the hint, let $\nu: A\to P_E$ be the function $a\mapsto [a]$.
Then
\[
\begin{array}{rl}
(a,b)\in E &\Longleftrightarrow [a]=[b]\\
&\Longleftrightarrow \nu(a)=\nu(b)\\
&\Longleftrightarrow (a,b)\in\ker(\nu).\\
\end{array}
\]
Since the sets
$E$ and $\ker(\nu)$ contain the same elements, they are equal.
\color{black}

\vfill

\begin{enumerate}
\item[3.]
  Suppose that $f:A\to B$ and
  $g:A\to C$ are two functions with common domain $A$.
  Let $f\times g: A\to B\times C$ be the
  product function: $a\mapsto (f(a),g(a))$.
  Show that $\ker(f\times g)=\ker(f)\cap \ker(g)$. 
\end{enumerate}


\bigskip

\color{blue}
\[
\begin{array}{rl}
(a,b)\in \ker(f\times g) &\Longleftrightarrow (f\times g)(a)=(f\times g)(b)\\
&\Longleftrightarrow (f(a),g(a))=(f(b),g(b))\\
&\Longleftrightarrow f(a)=f(b)\;\textrm{and} \;g(a)=g(b)\\
&\Longleftrightarrow (a,b)\in\ker(f)\cap \ker(g).\\
\end{array}
\]
\color{black}

\vfill  


\end{document}
