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Unformatted text preview: 15. OCTOBER 20TH: CANONICAL DECOMPOSITION 37 15. October 20th: Canonical decomposition 15.1 . Let f : A B , : A A/ , : im f B , f : A/ im f be as in this section. Since is injective, it has a leftinverse ; since is surjective, it has a rightinverse . Find a way to express f in terms of f , , . Answer: We have = 1 im f and = 1 A . Therefore f = 1 im f f 1 A/ = ( ) f ( ) = ( f ) = f . The last equality holds because f = f ; this was the whole point of the con struction of f (Theorem 15.1). Thus f = f . The reason why I did not use this approach in class is that the result, f = f , would seem to depend on the choices of the leftinverse of and of the rightinverse of . (Remember that leftinverses and rightinverses are not unique in general!) It does turn out that, magically, the function...
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This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.
 Fall '11
 Aluffi

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