A doubt in the proof of Kuratowski theorem In trying to understand the proof of Kuratowski's theorem (namely, a graph is planar if and only if it contains no subdivision of $K_5$ or $K_{3,3}$) from this book (Page 299) I am first trying to understand the proof of the fact that a minimal non planar graph where each vertex is of degree at least $3$ is $3$-connected. 
The book's proof is on the following lines. We start by noting that $G$ is $2$-connected. By way of contradiction, we then assume that $G=G_1\cup G_2$ with $V(G_1)\cap V(G_2)=\{x,y\},|V(G_i)|\ge 3$. Let $P_i$ be an $(x,y)$ path in $G_i$ and $H_i=G_i+P_{3-i}$. Then $H_i$ is planar and we can embed $H_i$ into the plane so that the path $P_{3-i}$ is on the boundary of the unbounded domain (this can be achieved by inverting the plane with respect to an appropriate circle). (And then the proof continues.)
I do not understand the last statement "we can embed $H_i$ into the plane so that the path $P_{3-i}$ is on the boundary of the unbounded domain". Can someone explain why is this so? I asked a more general question in this regard here but the answer to that does not solve the problem here. What is meant by "inverting the plane with respect to an appropriate circle"?
 A: Planar graphs, by definition, are graphs that can be represented (drawn) in the Euclidean plane using distinct points for vertices and mutually disjoint continuous paths between those points for edges joining them. If we drop the "mutually disjoint" condition then any finite graph can be represented in that way.
The definition would be an equivalent one if we replaced the Euclidean plane with the sphere. Stereographic projection from the North Pole onto the plane tangent to the South Pole establishes the necessary homeomorphism between the sphere (minus one point, the North Pole) and the Euclidean plane. Points close to the North Pole are projected onto points far away from the origin of the Euclidean plane and far away from the ensuing Euclidean representation of the graph.
But the sphere can be rotated with respect to the graph so that the North Pole actually lies inside one of the inner loops. In fact for any given edge of the graph there exist rotations of the sphere that place the North Pole in a region adjacent to that edge.
Applying stereographic projection to the new situation ensures that the given edge appears on the outside of the new Euclidean representation of the graph.
My guess is that the author means by "circle" any bounded connected component of the complement of the graph representation. I have called this an "inner loop" but it is the same thing. When the author says "unbounded domain" they refer to the unique unbounded connected component of the complement of the graph representation.
Now $P_{3-i}$ in the proof is a path, not an edge. But its topological role in $H_i$ is the same as if it were a single edge, because $P_{3-i}\cap G_i=\{x,y\},$ i.e., none of the intermediate points on the path has any edges (in $H_i$) apart from the two edges that make it part of the path. Therefore any circle of $H_i$ that is adjacent to a single edge of $P_{3-i}$ is adjacent to all edges of $P_{3-i}$ and consequently the above construction that rotates the North Pole next to an edge of $P_{3-i}$ ensures that all edges of the path $P_{3-i}$ are on the outside of the new Euclidean representation.
