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I am in the process of convincing myself of certain results:

I see that in the compact support cohomology mayer vietoris has opposite rows compared to the one in singular topology however it is said that those two cohomologies are the same in case the space is say a compact closed connected manifold. How is that?

I know I am very confused :) I hope I am not confusing you either ;)

Thank you in advance

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Let me do it with de Rham cohomology, for variety:

One defines the de Rham cohomology $H^\bullet(M)$ of a manifold $M$ starting from differential forms on $M$, and one defines the compactly supported de Rham cohomology $H_c^\bullet(M)$ of a manifold $M$ starting from differential forms on $M$ which have compact support.

Now, if $M$ is a compact manifold, every differential form on $M$ has compact support (because the support of such a thing is a closed subset of $M$, which is automatically compact in this case) so in this case the qualification «with compact suppot» does not, in fact, exclude any form. It follows immediately that $H^\bullet(M)$ and $H^\bullet_c(M)$ coincide.

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    Ok so by some (rational) magic although in both cohomologies Mayer vietoris does not seem the same it gives us the same results right?2011-05-21
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    @El Moro: I don't know what you mean by "Mayer vietoris does not seem the same".2011-05-21
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    Thank you for your comments. look at that http://www.srcf.ucam.org/~wjm29/Algebraic%20Topology%20Part%20III%20notes.pdf page 53 Theorem 9.122011-05-21
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    @El Moro: well, when $M$ is compact you have *two* M-V sequences. There is nothing wrong with that.2011-05-21
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    Check out Bott & Tu's "Differential Forms in Algebraic Topology", they do explicitly the case of $S^1$.2011-05-21