Let $\mathcal{A}$ and $\mathcal{B}$ be two abelian categories. Let $\tau : \mathcal{A} \rightarrow \mathcal{B}$ be a left-exact functor. Let $\mathcal{C}$ be the mapping cylinder category of $\tau$. Then, there are various functors $j_*, i_*$ ... defined on page 524 of Mazur's expository article in 1973, in Ann. Scient. ENS. He insisted that
the derived functor of $j_*$ are given by $$ R^q j_* M=(R^q \tau M, 0, 0) $$
(at l.8, page 525). Can someone explain this?
Form an injective resolution $$ 0 \to M \to I^0 \to I^1 \to I^2 \to \dots $$ and apply $j_*$. Then $j_* I^k = (\tau I^k, I^k, id_{\tau I^k})$ in the mapping cylinder category for all $k$, and so to calculate the derived functors you calculate the cohomology of the sequence $$ 0 \to (\tau I^0, I^0, id) \to (\tau I^1, I^1, id) \to \dots $$ Taking kernels or cokernels of a map $(N, M, \varphi) \to (N', M', \varphi')$ in the mapping cylinder category is done by applying it to both $N \to N'$ and $M \to M'$ individually, so to calculate the cohomology of the sequence at hand we take the cohomology of $$ 0 \to \tau I^0 \to \tau I^1 \to \tau I^2 \to \dots $$ (which are the derived functors $R^q \tau M$, by definition) and the cohomology of $$ 0 \to I^0 \to I^1 \to I^2 \to \dots $$ (which is $M$ in degree 0 and 0 in other degrees because this was a resolution of $M$). Therefore, the cohomology of the sequence in question is $$ (\tau M, M, id), (R^1 \tau M, 0, 0), (R^2 \tau M, 0, 0), \dots $$ which recovers Mazur's statement.
