By Lech Górniewicz
This ebook is dedicated to the topological mounted aspect idea of multivalued mappings together with purposes to differential inclusions and mathematical economic climate. it's the first monograph facing the mounted aspect thought of multivalued mappings in metric ANR areas. even if the theoretical fabric used to be tendentiously chosen with admire to functions, the textual content is self-contained. present effects are awarded.
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Extra info for Topological Fixed Point Theory of Multivalued Mappings (Topological Fixed Point Theory and Its Applications)
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For a vector space E and any integer q, define a linear map Θq : HomQ (E) ⊗ E → Hom(E, E) by letting Θq (u ⊗ v)(v ) = (−1)q u(v ) · v and extend Θq to all HomQ (E) ⊗ E. for u ∈ HomQ (E), v, v ∈ E, 48 CHAPTER I. 3) Lemma. If the vector space E is finite-dimensional, then Θq is an isomorphism. Proof. Let v1 , . . , vn be a basis for vector space E and v1 , . . , vn the conjugate basis to v1 , . . , vn . Then every element a in HomQ (E) ⊗ E has the following form: n aij vi ⊗ vj . a= i,j=1 If Θq (a) = 0 then n n akj vk (vk ) · vj = (−1)q Θq (a)(vk ) = (−1)q j=1 akj · vj = 0 j=1 so, akj = 0 for all k, j, which implies that a = 0.
So s is a continuous map and hence it is a proximative retraction. 11) Proposition. 1) for every y ∈ U there exists exactly one point x = x(y) ∈ A such that: y − x = dist(y, A). 12) Proposition. If A is a compact C 2 -manifold with or without the boundary, then A ∈ PANR. If A is a compact C 2 -manifold without boundary then taking a tubular neighbourhood of A in Rn we are able to obtain that A ∈ PANR. If the boundary ∂M of M is not empty the situation is more difficult because we have two tubular neighbourhoods (for M and ∂M ) but still we can obtain that A ∈ PANR.
If f: Y → Y is a proper map, then f is closed. Proof. Let A ⊂ Y be a closed subset of Y . We have to prove that f(A) is closed in X. Consider the sequence {xn } ⊂ f(A) such that limn xn = x. It is sufficient to prove that x ∈ f(A). In order to show it let us consider the set K = {xn } ∪ {x}. Then K is a compact subset of X and consequently the set f −1 (K) is compact. For every n we choose yn ∈ A such that f(yn ) = xn . Then {yn } ⊂ f −1 (K) and hence we can assume, without loss of generality, that limn yn = y.