# Download e-book for kindle: Boundedly Controlled Topology: Foundations of Algebraic by Anderson D.R., Munkholm H.J.

By Anderson D.R., Munkholm H.J.

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Additional resources for Boundedly Controlled Topology: Foundations of Algebraic Topology and Simple Homotopy Theory

Example text

3. In t h e ease p E k 2 we hence p ~ mI. c. a substitution from p A k I that is p r o v e d p = ~p A ~ k I = m I, for minimal (~,k1,~1,~,~,mT). 8k 2 = ek2 (= m 2 ) . 2. = pos(llm) £ ~1" The latter by definition By virtue of since ~m 6 R, am = Bm = m. implies (2210) (2216) and pos(~,ml) (iv). By V n 6 M(ml) , pos(l,n) (2212), there exists (T,k2,~2,B,l,m2) , which 2. ~2 ~ <2 is analogously (2223) Lemma: = pos(llml) completes assume = pos(l~m) ~k 2 = Bk 2 follows. a substitution the proof of <2 ~ ~2" proved.

Be satisfied: => p ~ m 2. mI = ~ kI E ~ k 2 = m 2. p is m i n i m a l , p C ~ and p A kI p ~ m I. 1. Consider the case p A k 2. 2. 2. 3. In t h e ease p E k 2 we hence p ~ mI. c. a substitution from p A k I that is p r o v e d p = ~p A ~ k I = m I, for minimal (~,k1,~1,~,~,mT). 8k 2 = ek2 (= m 2 ) . 2. = pos(llm) £ ~1" The latter by definition By virtue of since ~m 6 R, am = Bm = m. implies (2210) (2216) and pos(~,ml) (iv). By V n 6 M(ml) , pos(l,n) (2212), there exists (T,k2,~2,B,l,m2) , which 2.

1. 1. We assume m 6 C(111,ki) and will Clearly, am 6 C(112). Consider the case m ~ k I. H e n c e showam 6 C(112,ki). 2. In 6 C(112,ki). case m A k I assume: 3 h r-am, a k I. It h F" k I, k I A n 2 follows: (see I. 2. that ak I k I in c o n t r a d i c t i o n am A mk I = k I and We~assume First s E C ( 1 1 2 , n 2) that s = at. 1. 2. If s A k I, a s s u m e h 6 C(111) (iii) ~ak and s = mr. s 6 C ( 1 1 2 , k I) 3h such (see r 6 C(111,ki). 6[D I] = D 2 a n d We have a) a n d b) Let to to show are m 6 C(111) pos(111,m) I.