# Get A calculus for branched spines of 3-manifolds PDF

By Francesco Costantino

We determine a calculus for branched spines of 3-manifolds by way of branched Matveev-Piergallini strikes and branched bubble-moves. We in brief point out a few of its attainable purposes within the examine and definition of State-Sum Quantum Invariants.

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The adherence of B, Adh B, is the set of its adherent points. Exercise 112. Show that Adh B = n XEB CI X. To do this prove that both sides of the equation are equal to the adherence of the filter generated by B . Exercise 113. Let (E, T) be a topological space, A a subset of E . Prove that a point x of E is adherent to A if and only if there is a filter F on E such that A E F and F converges to x. 36 Chapter 4 If x is adherent to A th en VT(x) U {A} gene rates a filter with the required property.

Exercise 80. Let (E ,T) be a separable topological space. Prove that if V is a T-open subset of E then (V ,Tv) is also separable. Exercise 81. Let E be an uncountable set, p a point of E and Tp the particular point topology on E determined by p. Let A = Cdp}. Prove that (E, T p ) is separable but that (A, (Tp)A) is not. Example 2. Let ((Ei ,Ti))iEI be a family of to pological spaces. Let E = 0 iEI E, and, for each index i in J, let 1I"i be the projection mapping from E to E i . The topology induced on E by the family of mappings (1I"i)iEI is called the product topology on E ; we denote it by 0 iEI i: The topological space (0 iEl e; 0 iElTi) is called the topological product of the family ((Ei ,Ti)) iEI.

Adamson, A General Topology Workbook © Birkhäuser Boston 1996 44 Chapter 5 To prove this result, show first that the canonical surjection TJ from E onto E / R is an open mapping (to do this let U be any T-open subset of E and prove that TJ+-(TJ-> (U)) = U) . Now let X = TJ(x) and Y = TJ(Y) be distinct points of E / R ; since X and Yare distinct, it follows that (x,y) f/. R and so Clr{x} :I Clr{y}; so there is a T-open set U containing one of x and y but not the other. Show that TJ->(U) contains the corresponding one of X and Y but not the other.