Unformatted text preview: I is ﬁnite. This is enough for our applications since we can always choose a ﬁnite covering subfamily. The proof of injectivity is similar to the proof of Lemma 5.11 and is left to the reader. Let us show the surjectivity. Let ( m i a n i i ) i ∈ I ∈ limind i ∈ I M a i for some m i ∈ M and n i ≥ . Again we may assume that all n i are equal to some n . Since for any i, j ∈ I ρ ij ( m i a n i ) = ρ ji ( m j a n j ) , we have ( a i a j ) r ( a n j m ia n i m j ) = 0 for some r ≥ . Let p i = m i a r i , k = r + n . Then m i ra n i = p i a k i , f k j p i = a k i p j ....
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 Spring '09
 wei
 Algebra, Geometry, Morphism, Isomorphism, Epimorphism

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