TITLE: Is there a planar graph with two inequivalent embeddings on the sphere?
QUESTION [5 upvotes]: A planar graph is a graph with an embedding in the plane.  This graph can be embedded in the plane in two ways, and they are not related by a homotopy of graphs, a homotopy which preserves the graph's structure (vertices, edges etc.) and does not involve lines crossing each other.

However, as subsets of $\mathbb{R}^2$, they are homotopic without having to drag edges across each other.
This graph can be embedded in the plane in two ways, and it is not possible to change one into the other using a homotopy that doesn't cross edges over each other.

However, if these embeddings are lifted to the sphere, they become equivalent.
It is possible to have two inequivalent embeddings of a graph on the sphere:

but these only differ by a reflection.
Question:  Two embeddings of a graph on the sphere are equivalent if there is a homotopy of topological spaces from one to the other during which the graph does not intersect itself, possibly followed by a reflection. Is there a planar graph with two inequivalent embeddings on the sphere?

REPLY [5 votes]: Consider
O-----O           O---O--O
|    /|           |   |
| O O |   versus  | O |
|/    |           |/  |
O-----O           O---O