Strang, Introd. to Linear Algebra, Section 8.2 (Graphs and Networks): How is the expression for Kirchoff's Law obtained from this incidence matrix? I am following along Chapter 8 "Applications", Section 8.2 "Graphs and Networks" from Gilbert Strang's books Introduction to Linear Algebra.
In this section he shows examples of incidence matrices of graphs. As usual with this book, certain parts of the writing are unclear. I will go through the reasoning in the part I don't understand to show my question.
He gives an example of an incidence matrix
$$A=\begin{bmatrix} -1&1&0&0\\ -1&0&1&0\\ 0&-1&1&0\\ -1&0&0&1\\ 0&-1&0&1\\ 0&0&-1&1\\  \end{bmatrix}$$
of the directed graph

Each column is a node in a graph, and each row is an edge. The unknowns $x=\langle x_1,x_2, x_3,x_4\rangle$ represent potentials at the nodes. $Ax$ gives potential differences that cause flows.
$$Ax=\begin{bmatrix} x_2-x_1\\ x_3-x_1\\ x_3-x_2\\ x_4-x_1\\ x_4-x_2\\ x_4-x_3 \end{bmatrix}\tag{1}$$

How can we tell if a particular vector $b$ is in the column space of
an incidence matrix?

He then says

First answer: Try to solve $Ax=b$. That misses all the insight. As
before, orthogonality gives a better answer. We are now coming to
Kirchhoff's two famous laws of circuit theory-the voltage law and
current law. Those are natural expressions of "laws" of linear
algebra. It is especially pleasant to see the key role of the left
nullspace.
Second answer: $Ax$ is the vector of differences in equation (1). If
we add differences around a closed loop in the graph, the cancellation
leaves zero. Around the big triangle formed by edges 1,3, -2 (the
arrow goes backward on edge 2) the differences cancel:
$$\text{Voltage Law}\ \ \ \ \ (x_2-x_1)+(x_3-x_2)-(x_3-x_1)=0$$
The components of $Ax$ add to zero around every loop.

Up to this point everything is fine. The issue is in the following part

When $b$ is in
the column space of $A$ it must obey the same law:
$$\text{ Kirchoff's Law}\ \ \ \ \ b_1+b_3-b_2=0$$
By testing each loop, we decide whether $b$ is in the column space.
$Ax=b$ can be solved exactly when the components of $b$ satisfy all
the same dependencies as the rows of $A$. Then elimination leads to
$0=0$, and $Ax=b$ is consistent.

Where does the expression for Kirchoff's law come from? I am not asking about the physics, I am asking about how it came about from the matrices we see here?
Here is what I have so far (now after writing up the above):
He is taking the specific case of the vector $y=\langle 1,-1,1,0,0,0\rangle$ which we use to compute
$$y^TA=\begin{bmatrix} 0&0&0&0 \end{bmatrix}$$
This $y$ represents edges $1, 3,$ and $2$ (the latter going from node $3$ to $2$, and this is a loop.
For $b=\langle b_1, b_2, b_3, b_4, b_5, b_6 \rangle$ we have
$$y^Tb=b_1-b_2+b_3$$
Recall that the question asked in the book was "when do we know $b$ is in the column space"?
Apparently, the rationale is that if $Ax=b$ is true then $y^Tb=b_1-b_2+b_3$ must equal zero.
I seem to be missing a step because I can't see exactly why the last statement has to be true.
Final Edit:
If $Ax=b$ then $y^TAx=y^Tb$ then $0=y^Tb=b_1-b_2+b_3$. I think this is the reason.
 A: First of all, yes: your perspective is correct. Note that Strang is emphasizing the importance of the left nullspace. In this case, the vector $y = (1,-1,1,0,0,0)$ is an example of such an element. Now, recall the relationship between the "four fundamental subspaces": we have $\mathcal N(A^T) = \mathcal C(A)^\perp$. So, in order for $b$ to be an element of the column space $\mathcal C(A)$, it must be orthogonal to all elements of the left-nullspace $\mathcal N(A^T)$ (such as $y$).
In fact, you correctly prove this last fact in your "Final edit". If you turn back in the textbook to Strang's section on the four fundamental subspaces, you'll find a very similar step used in Strang's proof.
To address the direct jump from the "voltage law" equation to the "Kirchhoff's law" equation, we could justify the step directly as follows. If $b$ is in the column-space of $A$, then there is some vector $x = (x_1,x_2,x_3,x_4)$ for which
$$
b = \pmatrix{b_1\\b_2\\b_3\\b_4\\b_5\\b_6} = \pmatrix{x_2-x_1\\ x_3-x_1\\ x_3-x_2\\ x_4-x_1\\ x_4-x_2\\ x_4-x_3}.
$$
It follows that
$$
b_1 + b_3 - b_2 = (x_2 - x_1) + (x_3 - x_2) - (x_3 - x_1),
$$
which is zero as we already established.
