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I do not fully understand the proof in Wikipedia, the first paragraph of this.

$$ \begin{cases} v − e + f = 2 \\ 2e \ge 3f \end{cases} $$

Firstly $2e \ge 3f$ means that a facet has at least 3 edges, at the same time one edge exists besides exactly 2 facets - I guess that's how it is, right?

The conclusion $v \le 2f − 4$ however is still unclear. They simply took a legit $f \le 2v − 4$ and exchanged the variables. How is that allowed?

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See – user7530 Feb 19 '13 at 17:35
up vote 2 down vote accepted

You left out the justification given for the resulting inequality: "by duality."

If you need help understanding what "by duality" means and why it gives the resulting inequality, see this link to dual graph. In particular, read the properties, and the summary:

"Because of the duality, any result involving counting faces and vertices can be dualized by exchanging them." []

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I'd love to come up with the same property for $\mathbb{R}^4$. Do you think that's possible? Given that the above applies only to vertices-facets... – Ranas Feb 19 '13 at 17:58
Over spaces, over 300 today! Nealy 30k :-)) – Babak S. Feb 19 '13 at 18:14
@Babak: I've "capped" on upvote points (until 6 hours from now). I meant "accept" as in an accepted answer...there is no limit on points obtained from accepted answers. – amWhy Feb 19 '13 at 18:18

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