Augmenting path of edge-disjoint paths Consider the following definition of $\zeta$-augmenting path.
Let $\zeta$ be a set of edge-disjoint $uv$-paths. A $uv$-path $(w_0,w_1,...,w_r)$, labelled so that $w_0=u,w_r=v$, is a $\zeta$-augmenting path if, whenever $w_{i-1}w_i$ is an edge of some path in $\zeta$, say $w_{i-1}w_i$ is in the path $Q\in\zeta$, $w_i$ is closer to $u$ in $Q$ than $w_{i-1}$ is.
Question. With this definition, take $w_0w_1$ in $Q\in\zeta$. Then, for $w_1$ to be closer to $u$ in $Q$ than $w_0$ (which is $u$), $w_1$ must be $u$. Where am I going wrong with my logic? Can you give an example to clarify this definition?
 A: There is nothing wrong with your logic; the proper conclusion is simply that if $(w_0, w_1, \dots, w_r)$ is a $\zeta$-augmenting path, then there can be no path $Q \in \zeta$ containing edge $w_0w_1$.
The definition of a $\zeta$-augmenting path is designed so that, whenever such a path exists, we can use it to turn $\zeta$ into a larger collection of edge-disjoint paths from $u$ to $v$. Consider the following two simple examples:

For simplicity, let's have $\zeta$ consist of only one path in both cases: the path $(u,x_1,x_2,x_3,x_4,x_5,v)$. To be clear, both graphs are simple graphs; if I've drawn an edge twice, it's only to make it clear that it belongs to two paths.
In the example on the left (in red), the path $P = (u,w_1,w_2,w_3,v)$ is a $\zeta$-augmenting path: whenever it uses edges of a path in $\zeta$, it follows them "backwards". (For example, the edge $w_1w_2$ is such a shared edge, and $w_2$ is closer to $u$ than $w_1$.) As a result, we can use it to create two disjoint paths from $u$ to $v$: $(u, x_1, x_2, v)$ and $(u, x_4, x_5, v)$.
In the example on the right (in blue), the path $P = (u,w_1,w_2,w_3,v)$ is not a $\zeta$-augmenting path, because it follows some edges of path in $\zeta$ "forwards". (For example, the edge $w_1w_2$ is a shared edge, and $w_1$ is closer to $u$ than $w_2$.) We cannot create two disjoint paths from $u$ to $v$ here: both paths want to take the middle edges $w_1w_2 = x_2x_3$ and $w_2w_3 = x_3x_4$, and there's no way around that.
