I'm reading Bondy/Murthy's Graph Theory:
He defines $x$ as the source and $y$ as the sink, reading a bit later in the chapter, he presents this definitions:
$$ f^{+}(v)-f^-(v) = \left\{ \begin{array}{l l} val(f) & \quad \text{if $v=x$}\\ 0 & \quad \text{if $v\in X\setminus \{x\}$} \end{array} \right.$$
I can understand that $f^{+}(v)-f^{-}(v)=0$ when $v \in X\setminus \{x\}$, it is because what comes in a vertex is equal to what goes out of it. But I can't understand why $f^{+}(x)-f^-(x)=val(f)$, that expression would be the value of the flows coming to the source minus the value of the flows leaving the source. I'm confused because there is not flow coming to the source, isn't it? Considering this, it would be $0-f^-(v)$ which would be a negative number, the problem is that I guess that the value of the flow is positive. The it would need to be rewritten as:
$$|f^{+}(v)-f^-(v)|$$
I've tried to suppose that the value of the incoming flows is infinite and that the value of the outgoing flows is finite $(\infty-f^-(v))$[The value of the incoming flow would be limited by the capacity of the outgoing flows], but that also doesn't seem right.
And I also tried to presume that $f^{+}(v)-f^-(v)$ is not a subtraction but a symbol to that function (pretty much in the same way that we say that $f(x)=x$.)
I guess I got it. $f^{+}(x)-f^-(x) $ is the value of the flows outgoing minus the value of the flows incoming, as there is no flow incoming to the source, $f^{+}(x)-f^-(x)=val(f)$. I thought that it would represent the value of the flows coming to $x$ minus the value of the flows outgoing $x$. The interpretation is as in the image to the left, I was interpreting it as the image in the right.
$\quad \quad\quad\quad$