k - connectedness Diestel Problem: $G$ is $k$ connected and $xy$ is an edge in $G$. Show that $G/xy$ is $k$ connected iff $G-\{x,y\}$ is  $k-1$ connected.
Here is my approach:
$\Leftarrow$) This direction seems easy. Since for any two vertices of $G-\{x,y\}$ there are $k-1$ independent paths (Menger's Theorem) and for the same vertices $G$ has $k$ independent paths, it means that for any two vertices in $G$ there is always one vertex disjoint path passing through the edge $xy$. And hence $G/xy$ has $k$ independent paths between any two vertices. Which shows that it is $k$ connected.
$\Rightarrow$) In this direction I was unable to proceed. Since $G-\{x,y\}$ has lesser vertices, it is definitely $\leq k$ connected. However showing that it is $k-1$ connected, is equivalent to show that if we pick any two arbitrary vertices, then one independent path always passes through $xy$ in $G$. However how can we show this from the fact that $G/xy$ is $k$ connected, since contraction of an edge does not remove any edge.
 A: Let $v_{xy}$ be the contracted vertex in $G / xy$.  The main point is that $(G /xy) - v_{xy}=G - \{x,y\}$.  
Suppose $G - \{x,y\}$ is not $(k-1)$-connected.  Then there exists a subset of at most $k-2$ vertices $X$ such that $(G-\{x,y\})-X$ is disconnected.  But, $(G-\{x,y\})-X=G/xy-(X \cup v_{xy})$.  Thus, $X \cup v_{xy}$ is a vertex cutset of size at most $k-1$ in $G /xy$, and so $G /xy$ is not $k$-connected.
Conversely, suppose $G /xy$ is not $k$-connected.  Thus, $G /xy$ has a vertex cutset $X$ of size at most $k-1$.  Note that $v_{xy} \in X$, else $X$ is a cutset in $G$.  But now $X \setminus \{v_{xy}\}$ is a cutset in $G - \{x,y\}$ of size at most $k-2$.
A: Hint: if a $k$-separator of $G$ contains both $x$ and $y$ we get a $k-1$-separator of $G/xy$.
So every $k$-separator of $G$ misses either $x$ or $y$.
A: Let $G$ be a $k$-connected graph and let $xy$ be an edge in $G$.
Assume $G/xy$ is $k$-connected. As $G/xy$ is $k$-connected, it has at least $k + 1$ vertices, so $G$ has at least $k + 2$ vertices and $G − {x, y}$ has at least $k$ vertices. Any separator of $G/xy$ contains at least $k$ vertices, at most one of which is $v_{xy}$, so any separator of $G/xy − v_{xy}$ has size at least $k − 1$. As $G/xy − v_{xy} = G − {x, y}$, we conclude $G − {x, y}$ is $(k − 1)$-connected.
Now assume $G − {x, y}$ is $(k − 1)-$connected. Let $X ⊆ G/xy$. Assume that $|X| < k$. If $v_{xy} ∈/ X$, then $X ⊆ G$ but $X$ is not a separator of $G$ as it has size less than $k$ and $G$ is $k$-connected. So $G − X$ is connected, and as contracting and edge cannot disconnect a graph, $G/xy − X$ is not disconnected. If $v_{xy}∈ X$, let $X'= X\{v_{xy}\}$. Then $|X′|⊆G−{x,y}$, but $|X|<k−1$ so $X′$ does not separate $G−{x,y}$. As $G/xy−X = G−{x,y}−X′$, $G/xy$ is also connected. In both cases, $G/xy−X$ is connected, so $|X|<k⇒X$ is not a separator of $G/xy,$ 
so by the contrapositive,
$X$ is a separator of $G/xy⇒|X|≥k.$
Finally, if $G − {x, y}$ is $(k − 1)$-connected, it has at least $k$ vertices, so $G$ has at least $k + 2$ vertices, so $G/xy$ has at least $k + 1$ vertices. We conclude $G/xy$ is $k-connected$.
