Started my pass. NOT

done yet.

poly-form.tex

@@ -219,17 +219,25 @@ we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{G}{\
 
 Consider graph $G_2$, constructed from an arbitrary graph $G_1$.  We wish to show that the number of 3-matchings in $G_2$ will always be the linear combination above, regardless of the construction of $G_1$.
 
+\AR{I did not make a pass on the above since it looks incomplete and does not seem to have changed for a while. Also it would be good to define $G_2$ and $G_3$ outside of the proofs below.}
+
+\AR{You should have a subsection on the mapping $f$ that talks about the map independent of matchings. See comments below on how to state such observations.}
+
 \subsection{Three Matchings}
 \subsubsection{$G_2$}
 
 \begin{proof}
-Denote $3_{match}$ as the set of 3-matchings in $G_2$.  Denote $SG$ as the set of subgraphs imposed on $G_1$.  Notate each edge in $G_2$ as $(e, b)$ such that $b \in \{0, 1\}$, where $b$ identifies either the first or second edge of the two path that replaced the original $G_1$ edge.  Let $f: 3_{match} \mapsto SG$ be a function that maps a distinct 3-matching in $G_2$ to its generating subgraph.  An arbitrary $M \in 3_{match}$ in $G_2$ is then denoted by $(e_1, b_1), (e_2, b_2), (e_3, b_3)$, and $f(M) = \{e_1, e_2, e_3\}$.  Note that $f(M)$ is a set, i.e., the distinct edges of the generating subgraph in $G_1$.
+Denote $3_{match}$ as the set of 3-matchings in $G_2$. \AR{I do not like the notation $3_{match}$: perhaps $\mathcal{M}_3$ would be better?} Denote $SG$ as the set of subgraphs imposed on $G_1$. \AR{The set of all edge-subgraphs of $G_1$ already has an existing notation-- $2^{E_1}$, i.e. the power set of the set of edges of $G_1$.}  Notate each edge in $G_2$ as $(e, b)$ such that $b \in \{0, 1\}$, where $b$ identifies either the first or second edge of the two path that replaced the original $G_1$ edge.  
+
+\AR{I would recommend that you make the following changes. First define $f$ as a function $f:\binom{E_2}{3}\to \binom{E_1}{\le 3}$. As an aside, for any set $S$, the notation $\binom{S}{t}$ and $\binom{S}{\le t}$ denote the set of all subsets of $S$ of size {\em exactly} $t$ and {\em at most} $t$ respectively. Then for any $S\in\binom{E_2}{3}$, define $f(S)$ generically as below. Then argue that $f$ is properly defined: i.e. it is a function and $|f(S)|\le 3$. Also you'll need the definition for $G_3$ as well. One thought: define $G_k$ for $k>1$ generically. And then define $f_k$ to do the mapping from $G_k$ to $G_1$. Again, as mentioned above these definitions of $f$ and its properties should be pulled out up front.}
+Let $f: 3_{match} \mapsto SG$ be a function that maps a distinct 3-matching in $G_2$ to its generating subgraph.  An arbitrary $M \in 3_{match}$ in $G_2$ is then denoted by $(e_1, b_1), (e_2, b_2), (e_3, b_3)$, and $f(M) = \{e_1, e_2, e_3\}$.  Note that $f(M)$ is a set, i.e., the distinct edges of the generating subgraph in $G_1$.
 
 \underline{f is a function}:
+\AR{I think I had mentioned this before. In either case let me state it in a way so that you do not forget :-) {\LARGE \bf DO NOT USE UNDERLINING FOR EMPHASIS IN LATEX.} {\Huge \bf NEVER.} Underlining is a vestige of times when papers would be typeset using a typewriter. There are {\bf much} better ways to do it in Latex. In the current case since you'll be bringing this part out of the proof you can make the claim below a lemma.}
 
 First, let us argue that $f$ is indeed a function.  To do this, first note that for $G_2$ to contain a 3-matching, it must be that the generating subgraph in $G_1$ has at least 3 distinct edges.  This is because, with only 2 distinct $G_1$ edges, there are only two subgraph patterns, i.e., 2 disjoint edges and two-path, both of which when transformed into $G_2$ will not have enough disjoint edges to create a 3-matching.
 
-Consider then a minimal $G_1$, i.e. a graph with at most 3 edges.  Choose an arbitrary 3-matching $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ in the generated $G_2$ graph.  Choose an arbitrary edge $(e_i, b_i)$ and remove its corresponding generating edge in $G_1$.  Notice that this 3-matching now disappears from $G_2$.  Now replace the removed edge in $G_1$ with a new edge, placing it anywhere other than its original position.  Note that, no matter where the replacement edge lies, it cannot be that it generates the same 3-matching that the original edge helped uniquely generate, since a one-to-one correspondence exists between $(e_i, b_i)$ and $e_i$.  Thus, since any $M \in 3_{match}$ cannot be generated from more than one subgraph in $G_1$, $f$ must be a function.
+Consider then a minimal $G_1$, i.e. a graph with at most 3 edges.  Choose an arbitrary 3-matching $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ in the generated $G_2$ graph.  Choose an arbitrary edge $(e_i, b_i)$ and remove its corresponding generating edge in $G_1$.  Notice that this 3-matching now disappears from $G_2$.  Now replace the removed edge in $G_1$ with a new edge, placing it anywhere other than its original position.  Note that, no matter where the replacement edge lies, it cannot be that it generates the same 3-matching that the original edge helped uniquely generate, since a one-to-one correspondence exists between $(e_i, b_i)$ and $e_i$.\AR{this argument is overtly complicated. The main part of the argument is basically the last line. The argument should go like the following. For any choice of $b\in\{0,1\}$, the map $(e,b)\mapsto e$ is a one-to-one map and this implies that $f$ is indeed a function. Of course you should do this argument for the general $f_k$.}  Thus, since any $M \in 3_{match}$ cannot be generated from more than one subgraph in $G_1$, $f$ must be a function.
 
 \underline{For any $M \in 3_{match}, | f(M) |\leq 3$}
 
@@ -239,6 +247,7 @@ Since any arbitrary 3-matching $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ has an image
 \underline{ $M'$ s.t. $f(M') = G'$}
 
 First, note that there are a fixed number of \textit{distinct} 3-edge subgraphs that can appear on a given $G_1$.  Second, these subgraphs themselves are fixed, with the number of edges, vertices, and intersections always being the same.  Third, since each possible subgraph from a finite set of subgraphs is fixed, it must then be the case that each possible subgraph will always generate the same number of 3-matchings in $G_2$.
+\AR{I do not see the point of stating the claim above as well as the para above. The claim follows as a consequence of the argument below so it does not need to ``re-stated" above.}
 
 Specifically, we have the following possible subgraphs that can be composed of three edges.
 \begin{itemize}
@@ -315,4 +324,4 @@ In a similar fashion, enumerate through the various subgraphs in $G_1$ with $\le
 \[\numocc{G_3}{\threepath} = \numocc{G_1}{\ed} + 2 \times \numocc{G_1}{\twopath}.\]
 
 \subsection{Triangle}
-The number of triangles in both $G_2$ and $G_3$ will always be $0$ for the simple fact that when we replace a single edge with $\geq 2$-path, the possibility of a triangle of single edge sides disappears, since the only way a single edged triangle could exist is if it existed in $G_1$ and then was passed to $G_2$ or $G_3$ without replacing each single edge with $\geq 2$-paths.
+The number of triangles in both $G_2$ and $G_3$ will always be $0$ for the simple fact that when we replace a single edge with $\geq 2$-path, the possibility of a triangle of single edge sides disappears, since the only way a single edged triangle could exist is if it existed in $G_1$ and then was passed to $G_2$ or $G_3$ without replacing each single edge with $\geq 2$-paths.