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Vk . For each j = 1, . . , n we see that n ajk · vk = f(vj ). Θq (a)(vj ) = (−1)2q k=1 So, f and Θq (a) agree on the basis for E, which implies that Θq is onto. 3) is completed. Deﬁne e: HomQ (E) ⊗ E → Q as the evaluation map e(u ⊗ v) = u(v) for u ∈ HomQ (E), v ∈ E. 4) Lemma. If E is a ﬁnite-dimensional vector space and f: E → E is a linear map then q e(Θ−1 q (f)) = (−1) tr(f). Proof. Take a basis v1 , . . , vn for E and write n f(vj ) = ajk vk for j = 1, . . , n. 3) we know that n q Θ−1 q (f) = (−1) amk (vm ⊗ vk ), m,k=1 11.

Let us remark that any compact convex subset A ⊂ Rn is a PANR-space; then as U we can take Rn . Below we shall list important properties of PANR-spaces. 10) Proposition. Let A ∈ PANR. Then there exists an ε > 0 such that Oε (A) ∈ PANR. Proof. Because A ⊂ Rn is a proximative absolute neighbourhood retract there exists an open neighbourhood U of A in Rn and a proximative retraction r: U → A. Since A is compact, there exists an ε > 0 such that O2ε (A) ⊂ U . Now, for the proof we deﬁne a proximative retraction: s: O2ε (A) → Oε (A) by putting:  if x ∈ Oε (A), x s(x) = x − r(x)  r(x) + ε · if x ∈ O2ε (A) \ Oε (A).

It implies that for every j there exists a point xn,j ∈ An such that d(xj , xn,j ) ≤ εn . We can assume, without loss of generality, that limj xn,j = un ∈ An . Now, we claim that {un } is a Cauchy sequence, because d(xl,m , xs,m) ≤ εl + εs so we obtain d(ul , us ) ≤ εl + εs . Since (X, d) is a complete space we can assume that limn un = u. Now, it suﬃces to prove that the sequence {xn } contains a subsequence {xnj } such that lim xnj = u. Assume to the contrary that there exists δ > 0 such that d(xn , u) ≥ δ for every n.