Planar Graph & block

Question

a) Show that a graph is planar if and only if each of its blocks (maximal 2-connected subgraphs) is planar.
b) Deduce that a minimal nonplanar graph is a simple block.

Answer 1

a) It is clear that if G is planar, then each of its blocks must also be planar, as we can give a drawing of G, which must necessarily draw each block. It thus suffices to prove that if all blocks are planar, G is planar. We first note that two blocks have at most a size 1 intersection. This is because if they share two vertices, then the union of the two blocks must be two-connected, contradicting maximality (if you remove either of those two vertices, we can simply go through the other, so this is clearly two-connected).

Now that we have that these blocks intersect at at most one vertex, we can give a drawing of G. We go by induction over the number of blocks. Base case, of course, being G is a block, and is trivial. In the inductive step, we seek to remove a single block which intersects with only one other. This must exist, as otherwise every block is in a cycle of blocks. It is easy to tell that if we have a cycle of blocks, then they were all actually just one block in the first place, by a similar argument to above. Thus, we remove a single block except its intersection. By induction, the resulting G' is planar. Now, we draw this final block with the intersection point on the outer face, and similarly G' with the same intersection point on the outer face (this is always possible). Join them together, and we are done.

b) Take a minimal nonplanar graph. If it has more than one block, by part a, one must be nonplanar. But this is a smaller nonplanar graph, a contradiction.