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In Cauchy theorem and also in some Alexandrov theorems I came across the words "If $P,P'$ are Combinatorially Equivalent Polyhedra",i.e those that have "same structure". Does that mean the following?:

There is a 1-1 and onto correspondece $f:V\to V'$, where $V, V'$ are the vertices of $P,P'$ and $v_1v_2...v_n$ is a face of $P$, if and only if $f(v_1)f(v_2)...f(v_n)$ is a face of $P'$ .

Why just isomorpic graphs of $P,P'$ is not enough here for the polyhedra to have the same structure?

Thanks.

dmtri
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    For convex polyhedra, isomorphism of edge graphs does suffice, but this does not hold in higher dimensions - see e.g. this MSE thread or page 2 of this paper. I'm not sure whether it is enough for general (possibly nonconvex) polyhedra - off the top of my head, I can't think of any counterexamples. – RavenclawPrefect Jul 29 '21 at 21:53
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    Depends what you mean by same structure. Maybe it has something to do with orientation, like the difference between a "right handed" and "left handed" polyhedron. – TheManWhoNeverSleeps Jul 30 '21 at 04:54
  • @ TheManWhoNeverSleeps, the book says "the same structure", verbatim. I do not think orientation is an issue here. what I am asking more or less, here is: if 2 polyhedra (of $\Bbb{R^3}$) have isomorphic graphs through $f$, do their faces correspond through $f$. Thanks. – dmtri Jul 30 '21 at 06:26

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