Erik Demaine asks: << Can't any two topologically equivalent embeddings (in the sense that the cyclic order around each vertex is identical in two embeddings) be obtained by continuous deformation?

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This looks fairly easy to prove. And I'm fairly convinced of John's assertion that a straight-line embedding of a graph G in the plane *can* be obtained by a continuous deformation of an arbitrary planar embedding of G. But given that a graph G that can embedded in the plane can be embedded with all edges straight, however, doesn't immediately imply that the straight-line embedding *is* topologically equivalent to the original embedding (even though the graph itself is of course topologically the same). For example consider a square with a 5th node making a triangle with one side of the square -- a graph with 5 nodes and 6 edges. This can be embedded in the plane with the 5th node either inside or outside the square. (Of course, straightenability is not an issue here.) --Dan