Prove that flow is a linear combination of flow cycles and flow paths Let $D=(N,A)$ be a directed graph, and for an arc $e=xy$ define $h(e)=x$ and $t(e)=y$. A flow is $\mathbf{x}=(x(e_1),\dots,x(e_k))$ with $\sum_{e:t(e)=v}x(e)=\sum_
{e:h(e)=v}x(e)$ for all $v\in A\backslash \{s,t\}$. A flow cycle is a flow $\mathbf{y}$ on a directed cycle $C$ given by $y(e)=\epsilon(C)$ if $e\in E(C)$ and $y(e)=0$ otherwise, for some positive number $\epsilon(C)$. An $s-t$ flow path is a flow $\mathbf{z}$ on a directed $s-t$ path $P$ with $z(e)=\epsilon(P)$ for $e\in E(P)$ and $y(e)=0$ otherwise, for some positive number $\epsilon(P)$. Show that $\mathbf{x}$ can always be written as a sum of flow paths and flow cycles.
The first part of the problem was to show that a circulation, which is just a flow with no source or sink, can be written as a sum of flow cycles. I did this by way of the incidence matrix $A$ of $D$, showing that a circulation must be in the nullspace of $A$, which means we can form a basis of cycles, and since this basis is made of vectors with values either 0, -1 or 1, they are flow cycles. I can't figure out how to apply this approach in this case though since we are no longer dealing with the nullspace of the indicence matrix. Can the approach be adapted? Otherwise should I do something else? I'm really stuck here.
 A: First, your notation is a bit confusing, you have multiple $t$-s (tail and target) and multiple $x$-s and $y$-s (vertices and a flow), etc. (even using different fonts would help, like \mathrm and \mathbf, e.g. see the equations below).
As for the second part, the flow condition guarantees, that influx and outflux is the same for all vertices $v$ but for the source $s$ and target $t$
$$\sum_{e:\ \mathrm{t}(e)=v}\mathbf{x}(e)=\sum_{e:\ \mathrm{h}(e)=v}\mathbf{x}(e).$$
However, it also guarantees that the surplus of outflux in $s$ is the surplus of influx in $t$ (and given your definition also the other way around).
$$+\sum_{e:\ \mathrm{t}(e)=s}\mathbf{x}(e)-\sum_{e:\ \mathrm{h}(e)=s}\mathbf{x}(e) = 
-\sum_{e:\ \mathrm{t}(e)=t}\mathbf{x}(e)+\sum_{e:\ \mathrm{h}(e)=t}\mathbf{x}(e)
.$$
 So you can add another vertex $m$ that would connect $t \to s$, direct appropriate flow through it, then use the previous lemma to obtain cycle decomposition, and finally remove $m$ to break the cycle going through $m$ into $s-t$ path.
I hope this helps $\ddot\smile$
