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2 questions on coholomogy groups

by kakarotyjn
Tags: coholomogy, groups
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kakarotyjn
#1
Sep8-11, 08:33 AM
P: 97
Question 1: $$0 \to A\mathop \to \limits^f B\mathop \to \limits^g C \to 0$$ is an exact short sequence,in order to prove $$\cdots \to H^q (A)\mathop \to \limits^{f^* } H^q (B)\mathop \to \limits^{g^* } H^q (C)\mathop \to \limits^{d^* } H^{q + 1} (A) \to \cdots$$ is an exact long cohomology sequence,we need to prove$${\mathop{\rm Im}\nolimits} f^* = \ker g^* $$,I can prove $$
{\mathop{\rm Im}\nolimits} f^* \subset \ker g^* $$ because $${\mathop{\rm Im}\nolimits} f = \ker g$$,but how to prove $
{\mathop{\rm Im}\nolimits} f^* \supset \ker g^* $? \\
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Question 2: What makes the cohomology groups different by digging 2 points of $R^2$ from that of $R^2$ ?How does the closed and exact differential forms change?

Thank you!

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quasar987
#2
Sep8-11, 01:51 PM
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A1: Simple diagram chasing. Take [b] in Hq(B), b in Ker d such that g*[b]=[g(b)]=0. I.e. g(b)=dc for some c in Cq-1(C). To show: there exists a in Cq(A) and b' in Cq-1(B) such that b-db'=f(a).

Choose b' in g-1(c) (which exists since g is surjective). Note that since g is a chain map, g(b-db')=g(b)-dg(b')=dc-dc=0. So, since Ker(g)=Im(f), there exists a in Cq(A) such that f(a)=b-db'.

A2: This question is kinda vague, but notice that by digging points in R^2, new forms appear. Here is an example that illustrated the nature of the phenomenon: Consider on R-0 the 1-form (-ydx+xdy)/(x+y) [<---ill defined on all of R]. This form is closed but it is not exact. This is because it is equal to [itex]d\theta[/itex] everywhere where [itex]\theta[/itex] (the polar angle) is defined (i.e. R - {a half-ray}). Were (-ydx+xdy)/(x+y) exact, this would imply that there exists a differentiable extension of the polar angle function [itex]\theta(x,y)[/itex] to all of R-0. But clearly, there exists no such even continuous extension.
kakarotyjn
#3
Sep9-11, 09:02 AM
P: 97
Thank you very much! quasar987 :)


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