- #1
kakarotyjn
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Consider the infinite disjoint union [tex]M = \coprod\limits_{i = 1}^\infty {M_i }[/tex],where [tex]M_i 's[/tex] are all manifolds of finite type of the same dimension n.Then the de Rham cohomology is a direct product [tex]H^q (M) = \prod\limits_i {H^q (M_i )}[/tex](why?),but the compact cohomology is a direct sum [tex]H_c^q (M) = \mathop \oplus \limits_i H_c^q (M_i )[/tex](why?).
Taking the dual of the compact cohomology is a direct sum [tex]H^{q}_{c}(M)=\oplus_{i}H^{q}_{c}(M_i)[/tex](why?).
The question is indicated by red word,by the way I need to prove Kunneth Formula for compact cohomology using Poincare duality and the Kunneth formula for de Rham cohomology.But I don't know how to deal with the Dual space [tex](H^{n-q}_{c}(m))^{*}[/tex].Any ideas?
Thank you very much!
Taking the dual of the compact cohomology is a direct sum [tex]H^{q}_{c}(M)=\oplus_{i}H^{q}_{c}(M_i)[/tex](why?).
The question is indicated by red word,by the way I need to prove Kunneth Formula for compact cohomology using Poincare duality and the Kunneth formula for de Rham cohomology.But I don't know how to deal with the Dual space [tex](H^{n-q}_{c}(m))^{*}[/tex].Any ideas?
Thank you very much!
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