Dan, would you agree that any isormorphism of the given graph G would also be an isomorphism on its complement G-bar? G-bar just links all the integers into an infinite chain. And the proposed isomorphism is (to me) obviously impossible in G-bar. On Fri, May 15, 2020 at 8:41 PM Dan Asimov <dasimov@earthlink.net> wrote:
At the risk of sounding stupid, I'd like to see a proof of the claim (that the isomorphism of the subgraphs taking 1—>1, 2—>2, 4—>5, 5—>6 does not extend to an isomorphism G —> G).
Andy Latto wrote: -----
... the graph that links each integer n to every other integer except n-1 and n+1 would be a solution.
I don't think this works. The induced subgraph on {1,2,4,5} is isomorphic to the induced subgraph on {1,2,5,6}, using the isomorphism that fixes 1 and 2, sends 4 to 5 and 5 to 6. But this does not extend to an isomorphism of G to G. -----
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