Let $A$ be a unitary commutative ring, and let $B$ be an $A$-algebra. We consider the category whose objects are pairs $\textbf{M}=(M,f)$ where $M$ is an $A$-module and where $f$ is a $B$-linear endomorphism of $M\otimes\_A B$, and whose morphisms are $A$-linear maps whose scalar extension to $B$ make the obvious diagram commute, i.e. a morphism between $\textbf{M}$ and $(N,g)$ is a $A$-linear morphism $h:M\to N$ such that $g\circ (h\otimes\_A 1)=(h\otimes\_A 1)\circ f$. Let $\textbf{1}$ be the object $(A,\operatorname{id}\_B)$. I want to compute the extension group of $\textbf{1}$ by $(M,f)$ in the latter category. Hereafter, I sketched my computation:



> 
> The object $\textbf{P}=(A[X],X)$ is a projective object in this category, hence $0\to \textbf{P}\to \textbf{P}\to \textbf{1}\to 0$ where the first (nontrivial) arrow is multiplication by $1-X$ and the second one is the evaluation at $1$ is a projective resolution of $\textbf{1}$. 
> 
> 
> Let us compute $\operatorname{Hom}(\textbf{P},\textbf{M})$. We consider the $A$-linear morphism $\operatorname{Hom}(\textbf{P},\textbf{M})\to M$ which maps a morphism $h:A[X]\to M$ to its value $m=h(1)$. By commutativity of the "obvious" diagram $h(X^n)=f^n(m)$ which indicates that $f^n(m)$ belongs to $M$ for all positive $n$. Thanks to this remark, we easily obtain: 
>  $$\operatorname{Hom}(\textbf{P},\textbf{M})=\{m\in M~|~\forall n\geq 0:~f^n(m\otimes\_A 1)\in M\otimes\_A A\}=:f^{\bullet}(M).$$
>  Hence, $\operatorname{RHom}(\textbf{1},\textbf{M})$ is given by the complex $[f^{\bullet}(M)\to f^{\bullet}(M)]$ in degree $0$ and $1$ where the arrow is $1-f$.
> 
> 
> We conclude that $\operatorname{Ext^1(\textbf{1},\textbf{M})}$, which is given by the first cohomology group of this complex, is isomorphic to $\operatorname{coker}(1-f|f^{\bullet}(M))$. 
> 
> 
> 


However, let $m\in \operatorname{Hom}\_B(B,M\otimes\_A B)=M\otimes\_A B$. The object $(M\oplus A, \begin{pmatrix} f & m \\ 0 & 1 \end{pmatrix})$ is an extension of $\textbf{M}$ by $\textbf{1}$. Besides, it is equivalent to $(M\oplus A, \begin{pmatrix} f & m' \\ 0 & 1 \end{pmatrix})$ whenever 
$$\begin{pmatrix} f & m' \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & u \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & u \\ 0 & 1 \end{pmatrix}\begin{pmatrix} f & m \\ 0 & 1 \end{pmatrix}
$$
for some $u\in \operatorname{Hom}\_A(A,M)=M$, which boils down to $m'=m+u-f(u)$. Thanks to this argument, it seems that the extension group is rather given by
$$\operatorname{coker}(M\stackrel{1-f}{\longrightarrow} M\otimes\_A B).
$$
Can anyone point me out my mistake?
Many thanks!



Computation of extension groups in the category of pairs $(M,f)$
I have finally found an answer to my problem: in fact, the object $\textbf{P}$ is not a projective object in my category and hence, there is no clear projective resolution of $\textbf{1}$. The complex $\operatorname{RHom}(\textbf{1},\textbf{M})$, however, can still be computed using "rigid glued categories" as presented in Proposition 4.2.3 of  

*A. Huber, Mixed Motives and Their Realization in Derived Categories, Lecture Notes in Mathematics 1604*.
One easily find out that $\operatorname{RHom}(\textbf{1},\textbf{M})$ is naturally (in $\textbf{M}$) quasi-isomorphic to the complex of $A$-modules $[M\stackrel{1-f}{\longrightarrow} M\otimes\_A B]$ placed in degree $0$ and $1$, which explains my second computation.


