sorry if some of the terminology is wrong, since I translated this. Feel free to edit if you think I could state some of the terminology better or if I used wrong terminology.
I'm given the following version of Menger's theorem, and the corollary that is given next. I'm looking for a proof for the equivalence of the two. I have no idea where to start, so any help is much appreciated.
Menger's Theorem. The maximum number of pairwise arrow-disjoint $s-t$ paths is equal to the minimal cardinality of an $s-t$ cut.
Corollary. The maximum number of pairwise vertex-disjoint $s-t$ paths is equal to the minimal cardinality of an $s-t$ vertex cut.
Here $s-t$ cut is defined as the set of all arrows leaving $U$, with $s\in U$, $t\in V\setminus U$. The $s-t$ vertex cut is defined as a set $U$ such that $s,t\not\in U$ and such that every $s-t$ path has a vertex in $U$.
EDIT
I have thought about this question, and I might have an idea how to proceed. I'm am not sure if this approach is valid, though. So I'm still curious if anyone can think of a proof and if the proof below is valid.
So for the forward implication (Menger's theorem to the corollary), observe that all vertex-disjoint paths go through disjoint vertices, so an $s-t$ vertex cut for the maximum number of $k$ paths contains at least $k$ vertices, where $k$ equals the maximum amount of pairwise vertex-disjoint $s-t$ paths. Trivially there is an $s-t$ cut containing $k$ vertices; just choose from every one of the $k$ paths the second vertex (that is the first vertex after $s$). (i)
For the backward implication we look for a $s-t$ cut with minimal cardinality, so a set $U$ such that the set of arrows leaving $U$ is of minimal cardinality and so that $s\in U$, $t\not\in U$. We know that the vertex-disjoint paths go through $U$, trivially. Now we also look for arrow-disjoint paths, and I claim that the number of these paths equals the cardinality of the set of arrows leaving $U$. Assume that this is not the case. Than two paths go through the same arrow at some point, but leave $U$ through two arrows. Enlarging $U$ (or making $U$ smaller) by including the tail of the arrow instead of the two tails of arrows that the two paths have not in common, we obtain a set $U'$ such that the set of arrows leaving $U'$ is strictly smaller than the one of $U$. This is a contradiction, because I assumed that the cardinality of the set of arrows leaving $U$ was minimal. Thus there is a one-to-one correspondence between the vertices in $U$ and the disjoint $s-t$ paths, which proofs Menger's theorem.(ii)
Together (i) and (ii) prove the equivalence. $\qquad\blacksquare$
So actually no I have two questions: (1) is this proof valid? (2) Can you think of another and more direct or clearer proof?
I am really in doubt whether this proof is valid, because I have the feeling I don't use both versions of Menger's theorem enough for both implications.