$d$ operator for Mayer Vietoris sequence in De Rahm cohomology

Question

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.

Answer 1

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.