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\begin{document}

\begin{center}
{{\Large\bf ALGEBRA 1}\\
\bigskip

{\Large\bf HOMEWORK ASSIGNMENT I}\\

\bigskip

(Turn in underlined problems.)}

\bigskip

Read Chapter 0 and Section 1.1.
\end{center}

\vspace{.5in}

\noindent
\underline{\large\bf PROBLEMS}
\vspace{.25in}

{\bf 

\underline{1.} Let $X$ be a set, and let 
${\rm Inject}(X) = \lb\{f\in X^X\;|\;f\;\mbox{\rm is an
injection}\};\circ,\id\rb.$
Algebraize the theory of injective functions by showing that
\begin{enumerate}
\item[(a)] ${\rm Inject}(X)$ satisfies
\begin{enumerate}
\item[(i)] $f\circ (g\circ h) = (f\circ g)\circ h$,
\item[(ii)] $\id\circ f = f\circ\id = f$, and
\item[(iii)] $(f\circ g = f\circ h)\Rightarrow (g=h)$.
\end{enumerate}
(I.e.,  ${\rm Inject}(X)$ is a left cancellative monoid.)
\item[(b)] If $\m m = \lb M; \cdot, 1\rb$ is an algebra
satisfying (a)(i)--(iii), then $\m m$
is embeddable in  ${\rm Inject}(X)$ for some $X$.
\end{enumerate}

\bigskip

\underline{2.} Let 
${\rm Surject}(X) = \lb\{f\in X^X\;|\;f\;\mbox{\rm is a
surjection}\};\circ,\id\rb.$
Algebraize the theory of surjective functions by showing that
\begin{enumerate}
\item[(a)] ${\rm Surject}(X)$ is a right cancellative
monoid (i.e., satisfies 1(a)(i)-(ii) and 
\begin{enumerate}
\item[(iii)'] $(f\circ h = g\circ h)\Rightarrow (f=g)$.)
\end{enumerate}
\item[(b)] If $\m m = \lb M; \cdot, 1\rb$ is a
right cancellative monoid, then $\m m$
is embeddable in  ${\rm Surject}(X)$ for some $X$.
\end{enumerate}

\bigskip

Let ${\rm Biject}(X) = \lb\{f\in X^X\;|\;f\;\mbox{\rm is a
bijection}\};\circ,\id\rb.$
It would be natural to guess that 
${\rm Biject}(X)$ is algebraized by the class
of (left {\it and} right) cancellative monoids, but this is wrong.
Any algebra of the form 
${\rm Biject}(X)$ is indeed a cancellative monoid, but 
must satisfy
stronger properties that do not follow from cancellativity,
such as the one in the next problem.

\bigskip

\underline{3.} Show that ${\rm Biject}(X)$ satisfies
$$(f\circ g = f'\circ g')\&
(h\circ g = h'\circ g')\&
(h\circ k = h'\circ k')\Rightarrow
(f\circ k = f'\circ k').
$$

\bigskip

{\bf Challenge Problem!} (Prize = 100 Forint!) Find a cancellative
monoid that is not embeddable in ${\rm Biject}(X)$ 
because it fails the implication in Problem 3.

\bigskip

\underline{4.} Modify ${\rm Biject}(X)$ to 
${\rm Perm}(X) = \lb\{f\in X^X\;|\;f\;\mbox{\rm is a
bijection}\};\circ,{}^{-1},\id\rb.$
Here $f^{-1}$ is the inverse of the function
$f$. Show that 
the theory of bijective functions is
algebraized by structures 
$\m g = \lb G; \cdot, {}^{-1}, 1\rb$
satisfying 
\begin{enumerate}
\item[(i)] $x\cdot(y\cdot z) = (x\cdot y)\cdot z$,
\item[(ii)] $1\cdot x = x\cdot 1 = x$, and
\item[(iii)] $x\cdot x^{-1} = x^{-1}\cdot x = 1$.
\end{enumerate}
That is, show that 
${\rm Perm}(X)$ satisfies (i)-(iii), and that any structure
of type $\lb 2,1,0\rb$
satisfying (i)-(iii) is embeddable in
${\rm Perm}(X)$ for some $X$.

\bigskip

}
\end{document}

