G-v has k connected components

Question

Show that if G is a connected graph and every vertex has even degree, gr(v) = 2k and v is the only cut-vertex of G, then G-v has k connected components.

Answer 1

Notice $G-v$ is a graph with exactly $2k$ vertices of odd degree. Each connected component of $G-v$ has at least one vertex of odd degree (since it must have a neighbour of $v$ in the original graph). Each connected component must have an even number of vertices of odd degree (because the sum of all its degrees is even). It follows each connected component of $G-v$ has at least two vertices of odd degree, and therefore there can be at most $k$ such connected components.

Equality is not always achieved. Consider the following construction:

Take two copies of $K_7$ and for each of the copies select four vertices and remove all edges between these four vertices. Now add a new vertex $x$ and connect it only to the $4$ vertices of each copy. In this graph every vertex has even degree, the degree of $x$ is $8$ and the only cut vertex is $x$. However $G-x$ has only two connected components, not $4$.