Let $\Gamma$ be a digraph, $k$ a natural number and $a$, $b$ vertices in $\Gamma$ such that $$ \operatorname{outdegree}(a)-\operatorname{indegree}(a)=\operatorname{indegree}(b)-\operatorname{outdegree}(b)=k >0$$ and $$\operatorname{outdegree}(v)=\operatorname{indegree}(v)$$ for all other vertices in $\Gamma$ exept $a$ and $b$.
Show that there are $k$ directed walks from $a$ to $b$ in $\Gamma$ having no common directed edge pairwise.
I dont really know how to do this, but I was thinking if i could take some subgraph that is Euler graph and use that but im not sure if thats allowed or if there is any more simpler way
It is naturally to assume that the graph $\Gamma$ is finite. Start walk from $a$, not going along each edge twice. Since the graph $\Gamma$ is finite, we shall stick at some step. Since $\operatorname{outdegree}(v)=\operatorname{indegree}(v)$ for all vertices in $\Gamma$ except $a$ and $b$, we can stick only at the vertex $b$. So we went a walk from $a$ to $b$. It is easy to see that this walk contains a subwalk starting from $a$ and ending at $b$ without other occurrences of $a$ and $b$. Removing this subwalk from $\Gamma$, we see that the residual graph satisfies all the conditions stated for $\Gamma$, but with $k=k-1$ instead of $k$. So we can proceed the same way until we shall find $k$ pairwise edge-disjoint directed walks from $a$ to $b$.