Let $H$ be a quadrangulation on the sphere and let $f$ be a face of $H$ bounded by a $4$-cycle $∂f = v_1v_2v_3v_4$. A $4$-cycle addition to $f$ is to put a 4-cycle $u_1u_2u_3u_4$ inside $f$ and join $v_i$ and $u_i$ for $i = 1,\dots, 4$.
It is obvious that any quadrangulation is 2-connected.
My question is as follows:
If a quadrangulation $H$ is not 3-connected, then we apply a $4$-cycle addition to each face of $H$. Then, Does the resulting graph become $3$-connected?
It can be seen from the paper [A](Lemma 11) that the 2- vertex cut of a quadrangulation can only be two discontinuous vertices on the boundary of some 4-face. It seems that a 4-cycle addition can always break a 2-vertex cut locally, but this is just my intuitive view since connectivity of graph is a global graph property. I don't know how to explain it rigorously. For example, the following 4-cycle addition breaks the 2-vertex cut $\{u,v\}$.
Maybe we don't need to do 4-cycle addition to every 4-face and we only need to do 4-cycle addition to every face which boundary contains 2-cut vertex set of graph.
- [A] Noguchi K, Suzuki Y. Relationship Among Triangulations, Quadrangulations and Optimal $ $1 $$1-Planar Graphs[J]. Graphs and Combinatorics, 2015, 31(6): 1965-1972.
If a quadrangulation have a 2-cut, then there must a simple separating closed curve through a face for which $u$ and $v$ are diagonally opposite (this is the lemma you are referring to). If you apply your 4-cycle addition in this face, the closed curve can no longer pass through this face, because it doesn't exist anymore. If you apply a 4-cycle addition on every face where $u$ and $v$ are diagonally opposite, then clearly, $\{u,v\}$ can't be a 2-cut anymore. In fact you can even avoid one of these faces because the lemma state that the closed curve must pass through two such faces.
Now, we should verify that we have not introduced a new 2-cut. Clearly, there can't be a 2-cut using two old vertices because adding edges can't decrease the connectivity of the graph. Now lets take two vertex diagonally opposed in a face, where at least one is a new vertex. Lets consider the gadget formed by the face which was split in four faces. Lets take a face we created by adding a 4-cycle addition. There is two possibilities: one where we have two old vertices and two new vertices, and one where we have 4 new vertices. A 2-cut must use at least one of the newly introduced vertices. If we remove one of the old vertex, the graph must be still connected, because as you stated, every quadrangulation is clearly 2-connected. By a straightforward analysis, we see that no matter which 2-cut we choose, the corresponding gadget is still connected, and so the whole graph is still connected.
So your construction will yield a 3-connected graph.