Let $G$ be a group and suppose that there exists $d>0$ such that $H^s(G;M)=0$ for any $s>d$ and $\mathbb{Z}[G]$-module $M$. It is easy to see that the functor $M\mapsto H^d(G;M)$ is right exact. So we can define its left derived functor $\mathbb{L}_i H^d(G;-)=:L_i$.
I read the claim that this coincides with $H^{d-i}(M;)=:G_i$, but I am not sure about the proof. Since $L_i$ is a left derived homological functor it is universal, so we can lift the identity $Id\colon G_0\rightarrow L_0$ to a unique transformation of $\delta$-functors $G_i\rightarrow L_i$. Now we would like also $G_i$ to be universal so we can lift $Id\colon L_0\rightarrow G_0$ to another trasformation which must be the inverse of the first by uniqueness.
$G_i$ is not a derived functor and the classical theorem states that a homological $\delta$-functor is universal if it is effaceable (i.e. $G_i(P)=0$ for any $P$ projective module and $i\geq 1$). But I do not see why $G_i$ should have this property: $H^k(G;-)$ is zero only on injective modules, also for $i=s$ we get $G_s=(-)^{G}$ which is not necessarily zero on projective modules.
Is there any mistake in my reasoning or there exists a different proof?