By Gilbert Strang

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**Sample text**

Conversely, if D is a subcoalgebra, let i : D --~ C be the inclusion, which is an injective coalgebra map. Then i* : C* -~ D* is a surjective algebra map, and it is clear that Ker(i*) = ±. I t f ollows t hat D± i s a n i deal, a nd t he r equired i somorphism follows from the fundamental isomorphism theorem for algebras. A similar duality holds when we consider the ideals of an algebra and the subcoalgebras of the finite dual. 24 Let A be an algebra, and A° its finite dual. Then: i) If I is an ideal of A, it follows that ± nA°is a s ubcoalgebra in A°.

Ii) ¢1 is an isomorphism. iii) p is injective. If moreoverN is finite dimensional, then p is an isomorphism. Proof: i) Let x ¯ M*®Vwith ¢(x) = 0. Let z = ~if~®vi (finite sum), with fi E M*,vi ¯ V and (vi)i are linearly independent. Then 0 = ¢(x)(m) = ~ifi(m)vi for any m e M, whence f~(m) = 0 for and m. It follows that fi = 0 for any i, and then x -- 0. Thus ¢ is injective. Assumenow that V is finite dimensional. For V = k it is clear that ¢ is an isomorphism. Since the functors M* ® (-) and Horn(M,-) commute with finite direct sums, there exist isomorphisms n¢1 : M*® V --~ (M* ®k) and ¢~ : (Horn(M, k)) n -~ Horn(M, V), where n = dim(V).

Hence we can assume that f ¢ W± and so it follows that W ~ S±. Thus we can write W = (WClS ±)@W’, where We ¢ 0and dimk(W’) < oc. Also since f(S ±) = 0 and f(W) ¢ 0 it follows that f(W’) # O. , (n _> 1) be a basis for W’. Wedenote by ai f( ei) (1< i < n), hence not all the ai’s are zero. :(SnGeb ~#i Since ei ¢ S± @E key, then ei ¢ (S n r} e~-) ±. Hence there exists 9i e j¢i S Cl ~1 e~ such that gi(ei) = 1. So we have gi ~ S, and gi(ek) = fik. We denote by g = £ aigi. Hence g ~ S and g(e~) = a~ (1 < k < n).