T-Colouring of Hamiltonian Circuit in Cubic Graph I am trying to understand this paper On Hamiltonian Circuits.

I am unable to understand why the following is true :
$ X( S_{1}) +X( S_{2}) +X( S_{3}) = 0.$ $\: (1)$

Here $X(S)$ is representing the $1-chain$ on the cubic graph $mod\:2$ as coefficients. Its $1$ if $1-simplex \: \in S$ and  $0$ otherwise.
Now what is S? It is a S-subset, defined below.
Definition : It is a disjoint union of sets of simple closed curves of the cubic graph $N$, has an even number of $1-simplexes$ of $N$ s.t each $0-simplex$ of N is contained in some curve of set.
The number of components of $S-subset$ of $S$ is given by: $\sigma(S) + 1$

Definition :

$T-colouring$ of a cubical network $N$ is an unordered set of three
mutually exclusive classes, called $T-classes$, of $1-simplexes$ of
$N$ which together exhaust the $1-simplexes$ of $N$, and which are
such that each $0-simplex$ of $N$ is incident with one (and therefore
just one) member of each class.



THEOREM. In any cubical network $\sum_H X(H) = 0$, where $H$ runs
through the Hamiltonian circuits of $N$.

$Proof(Sketch)$: Here we want to associate $T-colouring$ with $S-subset$. Since the $T-colouring$ has three distinct mutually exclusive classes hence the equation $(1).$
I know that we want to cover all the $0-simplexes$ of $N$ and we separate them on the basis of what coefficient $1-simplex$ they belong to either $0$ or $1$. For $0$ its obvious and for $1$ its that they occur alternatively representing the respective T-colouring class.
Since our cubic graph has a Hamiltonian circuit it has $\sigma(S)=0$ and the number of distinct $T-colourings$ which can be associated with $S-subset$ is $2^{\sigma(S)}.$ So we can say :
$\sum_S 2^{\sigma(S)} X(S) = 0$ which is essentially equation (1), [lets write it again for convenience] : $ X( S_{1}) +X( S_{2}) +X( S_{3}) = 0.$
Again my question in this how the sum is equal to $0$? when we have assigned $1$ to the ($1-simplexes$ $\in$ $S-subset$ of S).
 A: We start with a $T$-coloring, which is just a proper $3$-edge-coloring of the graph. You can think of this as a decomposition of $G$ into three edge-disjoint spanning subgraphs: $G_1, G_2, G_3$. Each $G_i$ is a perfect matching, and together, they include every edge of $G$ exactly once.
If we take any union $G_i \cup G_j$, we get a $2$-regular spanning subgraph of $G$ (since every vertex is incident to one edge of $G_i$ and one edge of $G_j$). The only connected $2$-regular graphs are cycles, so $G_i \cup G_j$ is made up of cycles. Each cycle alternates between edges of $G_i$ and edges of $G_j$, so it has even length. This is exactly what Tutte is calling an $S$-subset.
Given the coloring $\{G_1, G_2, G_3\}$, we can form three $S$-subsets in this way: $G_1 \cup G_2$, $G_1 \cup G_3$, and $G_2 \cup G_3$. These are exactly the "associated $S$-subsets $S_1, S_2, S_3$" in the paper. We get
$$
   X(G_1 \cup G_2) + X(G_1 \cup G_3) + X(G_2 \cup G_3) = 0
$$
because an edge of $G_1$ has coefficient $1+1+0 = 0$ in this expression, an edge of $G_2$ has coefficient $1+0+1=0$, and an edge of $G_3$ has coefficient $0 + 1 + 1 = 0$.
