Dr. Katz DEq Homework Solutions 10

Dr. Katz DEq Homework Solutions 10 - 10 N.M. Katz: w hich...

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10 N.M. Katz: which is the negative of the isomorphism (1.2.1.8) explicated in (1.2.1.9), and which we will not use. In the setting of (1.2.1.9), we may choose over each V~ a morphism ~: ~1Vi~ Ker([3)[ V/~-fg[ Vi which is a section of ~: fq~--~ouf (i.e., such that ~ki. ~ = id,lv, ). (1.2.1.10.2) In fact, we may simply choose ~bi so that ~ o ~bi = id~rlv, - q~i ~ [3, this being possible because, over V~, id~e-Cp~o[3 is a projection onto ker([3). Then the difference ~bi- ~b~ defines a morphism from ~r V/~ V~ to f~l V~ n ~ which vanishes on the image of ct, thus defining by passage to quotients a morphism from ~1 V/c~ V~ to fq[ V/c~ V~, still denoted ~i- ~b~. The cohomology class of {~k~-~b~} in Ha(X, Hom(~, if)) is the element corresponding to the extension class of (1.2.1.1) via the isomorphism (1.2.1.10.1). To see that this isomorphism is the negative of (1.2.1.9), it suffices to recall that (1.2.1.10.3) 0~ o ~i = id~elv, - ~o~o [3 whence (1.2.1.10.4) ~ o (~k,- ~s) = - (~o,- ~os) o [3 which gives the equality of the two cocycles {~b i- ~bs} and {- (q~-cp~)}. (1.2.1.11) The "second" isomorphism (1.2.1.10) is the
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This note was uploaded on 12/21/2011 for the course MAP 4341 taught by Professor Normankatz during the Fall '11 term at UNF.

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