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Unformatted text preview: 15. OCTOBER 20TH: CANONICAL DECOMPOSITION 37 15. October 20th: Canonical decomposition 15.1 . Let f : A → B , π : A → A/ ∼ , ι : im f → B , f : A/ ∼ → im f be as in this section. Since ι is injective, it has a leftinverse κ ; since π is surjective, it has a rightinverse ρ . Find a way to express f in terms of f , κ , ρ . Answer: We have κ ◦ ι = 1 im f and π ◦ ρ = 1 A . Therefore f = 1 im f ◦ f ◦ 1 A/ ∼ = ( κ ◦ ι ) ◦ f ◦ ( π ◦ ρ ) = κ ◦ ( ι ◦ f ◦ π ) ◦ ρ = κ ◦ f ◦ ρ . The last equality holds because f = ι ◦ f ◦ π ; this was the whole point of the con struction of f (Theorem 15.1). Thus f = κ ◦ f ◦ ρ . The reason why I did not use this approach in class is that the result, f = κ ◦ f ◦ ρ , would seem to depend on the choices of the leftinverse κ of ι and of the rightinverse ρ of π . (Remember that leftinverses and rightinverses are not unique in general!) It does turn out that, magically, the function...
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 Fall '11
 Aluffi
 Equivalence relation, Binary relation, Bijection

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