By J. L. Dupont, I. H. Madsen

Collage of Aarhus, 50. Anniversary, eleven September 1978

**Read or Download Algebraic Topology, Aarhus 1978: Proceedings of a Symposium held at Aarhus, Denmark, August 7-12, 1978 PDF**

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**Additional resources for Algebraic Topology, Aarhus 1978: Proceedings of a Symposium held at Aarhus, Denmark, August 7-12, 1978**

**Sample text**

F / lies in a linear space which is called the span P of F and defined to be all points x 2 X that can be written in the form x D nj D1 aj xj , for ai 2 R, xj 2 F . F / is a finite-dimensional normed linear space, of dimension at most n. F / linearly homeomorphic to a euclidean space. 2. F / as a closed and bounded subset of a euclidean space and therefore it’s compact. F /. 1. F / is contained in C . F / is the intersection of all convex subsets of X containing F . Proof. Using induction on the number of points in F , the lemma is trivial for one point and we assume it is true for sets P of n 1 points.

D 0 or y 0 . / D 0. e A 1/ and the argument is pretty much the same in each case. s/ > 0 on . ; / and y 0 . / D 0. See Fig. 1. s//2 C B Fig. s/ > 0 and y. s/: This implies the corresponding relationship when we integrate Z s Z 2Ay 0 . /y 00 . y 0 . //2 C B 2Ay 0 . y 0 . //2 C B/ˇˇ ˇ ˇ 2Ay. /ˇˇ : s s Since y 0 . y. y. s/j Ä M for all s implies y. s/ Ä 2M . e A 1/: The final step of the proof is the easiest. s; u; p/j < M2 . 1 has completed the proof that S satisfies the hypotheses of the Leray–Schauder Alternative and consequently has a fixed point.

0/. Call a subset A of U admissible if A is compact and A \ @U D ;, where @U D U U is the boundary of U . We will require as part of the setting of the Brouwer degree that F is an admissible subset of U . Think of S n as Rn [ 1 so that U is a subset of S n . S n / ! U; U F/ F /. The excision property of homology F / ! S n ; S n F/ induces an isomorphism of homology. U; U F / by setting j 1 k . S n / is that generator that we just chose so carefully. U; U F /, but we can be sure that 0n is nontrivial, provided only that F is nonempty, for the following reason.