Definition of 2-connectedness in multigraphs While trying to prove that the dual of a 2-connected graph is 2-connected, I came across this proof in which something didn't make sense
http://people.math.sc.edu/lu/teaching/2013fall_776/homework6_sol.pdf
(Please refer to Question 3 in the link)
The part that didn't make sense was that in the base case, they considered the dual of the graph that was the smallest cycle. The dual of that graph had 2 vertices with multiedges. They said that this was 2-connected. However, going by the traditional definition of 2-connectedness (A graph is 2-connected if there doesn't exist a separating set of size 1), it isn't. 
So my question is, what is the definition of 2-connectedness in multigraphs
Thank You
 A: Another definition is given by Diestel in section 1.10 (fifth edition, page 28). Specifically:

The ends of loops and parellel edges in a multigraph $G$ are
  considered as separating that edge from the rest of $G$. The vertex
  $v$ of a loop $e$, therefore, is a cutvertex unless $(\{v\},\{e\})$ is
  a component of $G$, and $(\{v\},\{e\})$ is a 'block' in the sense of
  Chapter 3.1$^*$. Thus, a multigraph with a loop is never 2-connected, and
  any 3-connected multigraph is in fact a graph.

This definition is unintuitive, because it implies that a $K_4$ is 3-connected, but that adding a parallel edge to it would reduce its connectivity to 2-connected.
To capture this idea that an "edge" of a parallel graph can be disconnected from the rest, it is useful to imagine a vertex on it. If that fake vertex would be disconnected in a simple graph, the edge is disconnected in a multigraph. Here is a formal statement of that idea:

Consider a multigraph $G$. Create the simple graph $G^\prime$ by replacing each loop edge with a leaf vertex, and by adding a vertex bisecting each parallel edge. A vertex separator $X \subseteq V(G)$ separates $G$ if it also separates $G^\prime$. Note that separating vertices only come from the original graph.

* The definition of 'block' that he is referring to is a maximally connected subgraph of $G$ without a cut vertex of itself (although it may contain a cut vertex of $G$). I presume does not consider $v$ to be a cut vertex of $G = (\{v\}, \{e\})$ because there is nothing else for $e$ to connect to, just as deleting one vertex of $G = (\{x,y\}, \{xy\})$ does not disconnect it.
A: You are right to question this; indeed it is not obvious that the two-vertex graph with multi-edges is $2$-connected. There are two possible definitions of $k$-connectedness:


*

*The traditional definition is that a multigraph $G$ is $k$-connected if $|V(G)| > k$ and $G - S$ is connected whenever $S$ is a vertex set containing at most $k - 1$ elements.

*Alternatively, we could define a multigraph $G$ to be $k$-connected if $|V(G)| \geq \min(k,2)$ and any pair of vertices can be joined by at least $k$ independent paths (a set of path is independent if none of them contains an inner vertex of another).¹
For simple graphs, these two definitions are equivalent by Menger's theorem. However, the global vertex version of Menger's theorem (Diestel, theorem 3.3.6) no longer holds for multigraphs, so the above definitions are not equivalent in this case.
I'm not sure if there is consensus about this in the literature, but I would go with the first definition in the case of multigraphs. So then the graph from the solution you linked to is not 2-connected. However, it is $2$-connected in the sense that there are $2$ independent paths between any pair of vertices, and this might be what the solution author had in mind.
Alternatively, it might be that the solution author considers the complete graph $K_n$ to be $n$-connected. Indeed, for any set $S$ of at most $n - 1$ vertices, the graph $K_n - S$ is connected (and $K_n$ is the only $n$-vertex graph with this property). In light of the second definition above, it is usually declared that the connectivity of $K_n$ is $n - 1$ instead of $n$. One way to do this is by adding the requirement $|V(G)| > k$, as in the definition above (which is also Diestel's definition). Others solve this by considering a vertex set of $n - 1$ elements to be a vertex cut as well; see for instance this answer.
Still, this doesn't contradict the statement from the exercise. The course website shows that the exercises are taken from Diestel's Graph Theory, and Diestel is very precise in his definitions. Section 4.6 on plane duality is formulated in terms of multigraphs, but the problem at hand deals with a dual pair of plane graphs. In other words, it is assumed that both $G$ and $G^*$ are simple! So the given proof is not quite correct, but this doesn't contradict the exercise at hand.
References:


*

*Reinhard Diestel, Graph Theory, Fourth Edition, Graduate Texts in Mathematics 173, Springer, 2010.


¹: The condition $|V(G)| \geq \min(k,2)$ is only needed to exclude the trivial graphs on $0$ or $1$ vertices from being $k$-connected. Diestel omits this criterion and states in the first chapter that these trivial graphs will mostly be treated "with generous disregard". In other words: he basically assumes that any graph has at least two vertices to begin with.
