I am currently studying De Rahm cohomology, and as knowing manifolds was not a requirement for this class, we did everything on the open sets of $\mathbb{R}^n$.

I have a question for the Mayer-Vietoris sequence. We defined it as follows (I translated from french, so sorry if I use some wrong terminology).

Theorem. $\ $*Let $0 \xrightarrow{} A \xrightarrow{f} B \xrightarrow{g} C \to 0$ a short exact sequence of differential complexes, where $f$ and $g$ are chain applications. Then we obtain an exact sequence in cohomology $$\ldots \to H^{q-1}(C) \xrightarrow{d^\ast} H^q(A) \xrightarrow{f^\ast} H^q(B) \xrightarrow{g^\ast} H^q(C) \xrightarrow{d^\ast} H^{q+1}(A) \to \ldots$$ which is called the Mayer-Vietoris sequence.*

Now for explicit calculations, we considered an open set $M \subset \mathbb{R}^n$ which we write as $M = U \cup V$ for some open sets $U$ and $V$. I skip a bit of details, but we end up, by Mayer-Vietoris, with the following exact sequence: $$\ldots \to H^{q-1}(U \cap V) \xrightarrow{d^{q-1}} H^q(M) \xrightarrow{i^\ast} H^q(U) \oplus H^q(V) \xrightarrow{\partial^q} H^q(U \cap V) \xrightarrow{d^q} H^{q+1}(M) \to \ldots$$

My question is: what can we say about the $d^q$ operator for example? Is it surjective for example, or injective? I could not figure this out with the construction in the proof of Mayer-Vietoris.

Thanks for any answer.

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    $\begingroup$ In general, you can not say anything about the boundary operator per se. $\endgroup$ – iwriteonbananas Jan 17 '16 at 9:22
  • $\begingroup$ @iwriteonbananas I asked this, because I studied an example to compute the cohomology groups of the $2$-torus, and at some point, it is said "thus the surjective map $\delta$ going into $H^2(T^2)$ has kernel $\mathbb{R}$", and I did not understand where the surjectivity cam from. source: math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Greene.pdf page 11, Proposition 3.12 , $\endgroup$ – Laurent Hayez Jan 17 '16 at 9:45
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    $\begingroup$ Use exactness- in the case in question the term after the target of delta is zero. $\endgroup$ – Dylan Wilson Jan 18 '16 at 13:25

The answer is given in the comments. In general you can't say if a certain map is injective or surjective without further information. Knowing that certain groups are zero will show that certain maps are injective (the one after the map that has as a domain the zero group) or surjective the map before the one mapping to zero). It is also good to know that a long exact sequence can be broken down in short exact sequences. See for example here short exact sequence breakdown.

  • $\begingroup$ Hey thanks. I saw that when I computed the Mayer-Vietoris sequence myself. I thought it was stated as a general fact in the text, hence my confusion. Thank you for the precisions. $\endgroup$ – Laurent Hayez Jan 20 '16 at 8:28

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