There has to be some mistake in your reply :-) Here I am posting the drawing with three crossings with shown isomorphism.

The crossing number of Heawood graph is 3. The task is to prove it.

Yes, my bad. e = 21 for the graph, not 42. And the inequality does not deliver any useful estimate. Will have to think about it.

Another way to get the result that c >= e - 3v + 6 is to consider removing just one of the edges involved in each crossing. This leaves a planar graph, still on v vertices, but now with e' edges. At most c edges are removed. So c >= e - e' >= e - 3v + 6

However, we can go much further in this analysis with the Heawood graph, which has been carefully designed to contain no 3, 4 or 5-cycles:

The edges divide in an obvious way into long and short edges. Clearly there is no such cycle composed entirely of short edges, or with just 1 long edge. If there were 2 long edges, they could not be adjacent and so would both contribute +5 to the cycle. the remaining edges (at most 3) could neither complete this to a full loop (of length 14) or reduce it to 0. There could not be more than 2 long edges, as then some two would have to be adjacent.

So if it was planar, each face would have to be surrounded by 6 edges, and so as before we would have 6f <= 2e, so 3f <= e, so 6 = 3f + 3v - 3e <= 3v - 2e. This now shows that it is not planar. We cannot proceed, as alexb did, by adding a vertex at each crossing, inducing 2c new edges: The graph this produces may contain faces with fewer than 6 edges. Instead, we consider removing edges to make the graph planar. However, if we remove edges, we are still left with a graph with no cycles of length 3, 4 or 5. So we can still use the identity 2e' <= 3v - 6. As before, c >= e - e' >= e - (3v-6)/2 = 3 for the Heawood graph.

Thankyou

sfwc
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