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## Cutsets

Proof: This is an induction on . For n=0 there is nothing to prove. So now let n>0 and suppose the result holds for all k<n. Let and let , and let . Then K' and B' satisfy the hypothesis of the theorem: In fact if is a simplex in K', then by the clique condition is a simplex in K and again via the clique condition

where the latter simplicial complex is contractible. Moreover is contractible by assumption. Now since by induction hypothesis K' is contractible and thus x is escamotable.
To prove the ``moreover" part we need to prove that K is ``dismantlable via escamotable points" to K[B]. To see this it is good enough to show that and B satisfy the assumption of the theorem. Let be a simplex. Then is contractible by assumption and we are done. \

Clearly a set B as above must intersect every maximal simplex, which is exactly the notion of a cutset.

Proof: We will prove that the simplicial complex satisfies the assumption of Theorem 4.41. Let be a simplex. Then the set has a center and is thus contractible. \

Bernd.S.W.Schroder