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
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:
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.