Kdtyjb

Good day,
I am trying to end the subsection Proof on line ~65 and have the following section reset its indentation to be inline with the previous section, not the subsection Proof on line ~65. I will be doing this throughout my document so if you could not only troubleshoot this issue for me, but include instructions for subsequent areas I would be greatly appreciative.

Thank you for all your assistance,

jhayes

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usepackage{textcomp}

usepackage{indentfirst}



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newenvironment{tab}

  {begin{tabbing}

           hspace{30pt}=hspace{30pt}=kill

  }

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title{MATH 570 Lecture Notes}

author{Professor: Dr. Jon Corson\

Prepared by: Justin Hayes}



date{Spring 2019}



begin{document}



tableofcontents



maketitle



chapter{Review of Functions}

section{Definition}

indent A function $Phi$ from a set A (domain) to a set B (co-domain) [also known as mapping] [if from set A to set A then known as transformation] is a rule/relation that assigns each a$in$A a unique element $Phi$(a) in B.



vs

Two functions($Phi$ & $Psi$) are equal (denoted $Phi$=$Psi$) if they nave the same domain $&$ co-domain $&$ exactly the same rule and $Phi$(a)=$Psi$(a) for each a in the common domain.



section{Identity Function:}

For any set A, the identity function on A is the function I$_A$:A$rightarrow$A given by I$_A$(a)=a for all a$in$A.



section{Composition of Functions:}

If $alpha$:A$rightarrow$B and $beta$:B$rightarrow$C are functions then the composite of functions:\

$betacircalpha$=$betaalpha$:A$rightarrow$C is defined by ($betaalpha$)(a)=$beta$($alpha$(a)).



section{Associative Law:}

If $alpha$:A$rightarrow$B, $beta$:B$rightarrow$C; $gamma$:C$rightarrow$D,\

then $gamma$($betaalpha$)=($gammabeta$)$alpha$



subsection{Proof:}



begin{center}

    Both composites have domain $A$ & co-domain $D$.\

    For each element in the domain ($alphain$A),\

    $lbrackgamma$($betaalpharbrack$(a)$Rightarrowgamma$(($betaalpha$(a))=$gamma$(($beta$($alpha$(a)))\

    $lbrack$($gammabeta$)$alpharbrack$(a)$Rightarrow$($gammabeta$)($alpha$(a))=$gamma$($beta$($alpha$(a)))\

    Therefore $gamma$($betaalpha)$=($gammabeta$)$alpha$.

end{center}



section{Identity Law:}

If $alpha$:A$rightarrow$B, then\

I$_Bcircalpha$=$alpha$ and $alpha$ $circ$ I$_A$=$alpha$



section{One-to-One, Onto & Bijections}

subsection{One-to-One}

We say that $alpha:Arightarrow B$ is textbf{1-to-1}(injective) $iffalpha$(a$_1$)=$alpha$(a$_2$)$Rightarrow$(a$_1$)=(a$_2$)



subsection{Onto}

We say that $alpha$:A$rightarrow$B is textbf{onto} (surjective) if for each element in the codomain B $exists a in A$ such that $alpha$(a)=b.



subsection{Bijection}

We say that $alpha:Arightarrow B$ is textbf{bijective} if $alpha$ is both 1-to-1 textbf{&} onto.



subsection{Theorem of Bijection:} 

A map $alpha$:A$rightarrow$B is a bijection if and only if there exists a map $beta$:B$rightarrow$A such that $betaalpha$=I$_A$ and $alphabeta$=I$_B$.\

textbf{Remark:} In this event, the function $beta$ is unique.



subsection{Proof of Bijective Theorem:}

begin{center}

    Suppose B':B$rightarrow$A is another such function.\

    Then: B'=I$_Acirc$B'=$beta$($alpha$B')=$betacirc$I$_B$=$beta$\

    textbf{Notation:} we denote the unique function $beta$ in the theorem by a$^{-1}$, called the inverse of the bijection of $alpha$.\

    $alpha^{-1}alpha$=I$_A$ & $alphaalpha^{-1}$=I$_Bsquare$\

end{center}

From your graphic everything is aligned on the first page of the ToC and is also aligned on the second page; the difference is that the left margin on the two pages is different which you get by default with the book class. If you want the left margin to be constant throughout your document use the oneside option.

documentclass[oneside]{book}