# An Alpine Anthology of Homotopy Theory by Dominique Arlettaz

By Dominique Arlettaz

The second one Arolla convention on algebraic topology introduced jointly experts masking a variety of homotopy thought and $K$-theory. those court cases replicate either the range of talks given on the convention and the variety of promising study instructions in homotopy idea. The articles contained during this quantity comprise major contributions to classical volatile homotopy concept, version type concept, equivariant homotopy thought, and the homotopy concept of fusion platforms, in addition to to $K$-theory of either neighborhood fields and $C^*$-algebras.

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Additional resources for An Alpine Anthology of Homotopy Theory

Example text

Those produced in this way are all equivalent cobordisms in the sense we now make precise. 17 Equivalent cobordisms. Given two oriented cobordisms from 1, 0 ✲ M ✛ ✲ ✛ 0 to 1 M we say they are equivalent if there is an orientation-preserving diffeomorphism ∼ M making this diagram commute: ψ :M→ ✲ 0 ✲ M ✛ ✻ ψ 1 ✛ M (Note that the two triangles truly commute – not just up to diffeomorphism. ) In the next subsection we will divide out by these equivalences, and consider equivalence classes of cobordisms, called cobordism classes.

13, it is not at all obvious that the attachment does not provide new possible smooth structures. 12. Recall that every cobordism already decomposes into a cylinder followed by something else. Precisely, let M be our cobordism from 0 to 1 , and let C denote a cylinder over 0 . We want to show that up to diffeomorphism rel the boundary, CM = M. Decompose M as M = M[0,ε] M[ε,1] where the ﬁrst part is diffeomorphic to a cylinder over 0 . Now we can ﬁnish the proof by writing (modulo diffeomorphism): CM = C(M[0,ε] M[ε,1] ) = (CM[0,ε] )M[ε,1] = M[0,ε] M[ε,1] = M.

Vn ] (respectively [w1 , . . , wm ]) is a positive basis for Tx X (respectively Ty Y ), then [v1 , . . , vn , w1 , . . , wm ] is declared to be a positive basis for T(x,y) (X × Y ). , the reader may object. Well, yes and no: you get ‘another’ orientation because X × Y is not the same manifold as Y × X. Of course they are isomorphic, and ∼ Y × X, (x, y) → (y, x). the natural isomorphism is the twist map X × Y → Now if you compare the two orientations carefully along this isomorphism you will note that they agree!