Is the dual graph of a planar graph an isomorphism invariant?

Question

Is the dual graph of a planar graph an isomorphism invariant of the graph? Or does it depend on the embedding in $S^2$?

If it is an invariant, what's the proof?

Here's what I mean. Given a planar graph $\Gamma = (V,E)$, define an embedding of a planar graph to be a cell complex whose total space is homemorphic to $S^2$ and a map from $V$ to the $0$-cells and $E$ to the $1$-cells that respects the incidence structure. (An aside: note that this definition implies $\Gamma$ is connected. I know this isn't standard but bear with me.)

My understanding of the dual graph depends on such an embedding: I define $\Gamma^\vee$ to have as vertices the $2$-cells of the cell complex and as edges the $1$-cells, with an edge connecting the pair of vertices / $2$-cells to which it is incident.

My question would be answered either by a definition of the dual graph that didn't depend on such an embedding, or by a proof that there is an isomorphism between the duals resulting from any two embeddings. (Edit: Oh, right. Or, examples to show that the dual graph isn't an isomorphism invariant!)

Thanks in advance!

Wikipedia states that the duals corresponding to different embeddings can indeed be different, but does not give an example. Later. Ah. There is an example here — Mariano Suárez-Álvarez, Feb 04 '15 at 23:21
I had occasion to draw a picture of this in one of my answers (two planar embeddings of a graph that give distinct duals). — hardmath, Feb 04 '15 at 23:49

score 3 · Accepted Answer · answered Feb 05 '15 at 02:25

If a graph is 3-connected and planar, then it has a unique embedding on the sphere. This a result of Whitney (see the Wikipedia article on dual graphs). If the graph is not 3-connected then, as the examples in the comments show, the embedding is in general not unique.

Is the dual graph of a planar graph an isomorphism invariant?

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