\hfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section*{Problems and Comments For Section 3} \vspace{1pt} \textbf{Problems: }3.1, 3.4, 3.9, 3.10, 3.11, 3.12 \vspace{1pt} \textbf{Comments: }For the multiplicative monoid of $n\times n-$matrices one has that a left-inverse of a matrix $A$ is automatically a right-inverse, thus an inverse. \newline This is proven in linear algebra and related to the \begin{theorem} The linear homogeneous system $AX=0$ of $n$ equations in $n$ unknowns has only the trivial solution if and only if $AX=B$ has for every $B$ (exactly) one solution. Here $A$ is an $n\times n$ matrix and $X$ and $B$ are $n\times 1$ matrices. \end{theorem} For the monoid of maps on a set $X$ one has that the map $f:S\rightarrow S$ has a left inverse $g:S\rightarrow S,$that is $g\circ f=id_{S}$ for some $g, $if and only if $f$ is \textit{injective}, that is $f(x_{1})=f(x_{2})$ iff $% x_{1}=x_{2}.$And $f$ has a right inverse $h:S\rightarrow S,$ that is $f\circ h=id_{S}$ for some $h$, if and only if $f\ $ is surjective, that is for every% $\ y\in S$ \ there is some $x$ such that $f(x)=y.$ It now follows:: \begin{theorem} If a map $f:S\rightarrow S$ has a left as well a right inverse, then it has a unique inverse, which is the inverse map $f^{-1}$ of $f$% \[ f(x)=y\iff f^{-1}(y)=x \] \end{theorem} For maps on \textit{finite sets }$S$ one has that $f:S\rightarrow S$ is injective if and only $f$ is surjective. Why? \begin{exercise} Let $S=% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ and $f:n\longmapsto 2n$. Find a left inverse $g$ and demonstrate that it is not a right inverse and that there are many left inverses for $f.$ Now let $f:n\mapsto d(n)$, where $d(n)$ is the number of prime divisors of $n.$ Find a right inverse $g$ and demonstrate that it is not a left inverse and that there are many right inverses for $f.$ \end{exercise} \end{document}