I am trying to understand some proof of the fact that given a real vector space of dimension n, a ring $A$, some derive object $M$ from the derived category of $A-$modules bounded from below, then $R\Gamma_{c}(\mathbb{R}^{n},M)\simeq M[-n]$. The author assume at first that $n=1$, note $U_{1},U_{2}$ connected components of $\mathbb{R}^{*}$, and exhibit a distinguished triangle $R\Gamma_{\{0\}}(\mathbb{R},M)\longrightarrow R\Gamma(\mathbb{R},M) \longrightarrow R\Gamma(U_{1},M) \bigoplus R\Gamma(U_{2},M)\longrightarrow (+1) $
from the fact that $R\Gamma(\mathbb{R},M)\simeq M_{x}$, for any $x\in\mathbb{R}$ (from $\mathbb{R}$ contractibility), they conclude that the first morphism is null because the second morphism has a left inverse. But why the second morphism has a left inverse ?
for any complex $M\in D(A)$, we have $R\Gamma(\mathbb{R},M)=M$. Indeed, if $M$ is concentrated in degree $0$, then there is no higher cohomology and the global sections are the locally constant (hence constant) functions $\mathbb{R}\to M$. But this immediately implies that the same is true for any complex $M$.
Since $U_1, U_2$ are homeomorphic to $\mathbb{R}$, we also have canonical isomorphisms $R\Gamma(U_1,M)=M$ and $R\Gamma(U_2,M)=M$.
Moreover the induced map $R\Gamma(\mathbb{R},M)\to R\Gamma(U_1,M)\oplus R\Gamma(\mathbb{R},M)$ gives modulo the above isomorphisms the diagonal map $M\to M\oplus M$. Again, to see this, first assume that $M$ is concentrated in degree $0$, then remember that the map is given by restricting the constant function $m$ on $\mathbb{R}$ to $U_1$ and to $U_2$, so this is obviously the diagonal. Again, this immediately implies that this holds for a general complex $M$.
Finally, it is clear that the map $M\to M\oplus M$ has a left inverse, which proves the claim.