Let $\mathcal{C}$ be an additive category, $\Psi (C)$ the category of all complexes over $\mathcal{C}$ with chain maps, $\mathcal{K} (C)$ the homotopy category of $\Psi (C)$, the derived category of $\mathcal{C}$ denoted by $\mathcal{D} (C)$, $ \mathcal{D}(C)$ is the localization of $\mathcal{K} (C)$ at all quasi-isomorphisms.

Suppose $X,M,Y$ are objects in $\mathcal{C}$ $\mathcal{}$, we denote by $add(M)$ the full subcategory of $\mathcal{C}$ consisting of all direct summands of direct sums of finitely many copies of $M$. Let $M_{0} \in add(M)$, $ X \rightarrow M_0 \rightarrow Y$ be an $add(M)$-exact sequence in $\mathcal{C}$ .

Supose $V :=X \oplus M$, $\Lambda :=End_{\mathcal{C}} (V)$ Moreover, there exists the following exact sequence of $\Lambda$-modules: $0 \rightarrow Hom_{\mathcal{C}} (V,X) \stackrel{f}{\longrightarrow} Hom_{\mathcal{C}} (V,M_0) \stackrel{g}{\longrightarrow} Hom_{\mathcal{C}}(V,Y)$. Let $T :=Hom_{\mathcal{C}}(V,M) \oplus Im(g)$. Then we can get another exact sequence of $\Lambda$-modules: $0 \rightarrow Hom_{\mathcal{C}} (V,X) \rightarrow Hom_{\mathcal{C}} (V,M \oplus M_0) \rightarrow T \rightarrow 0$.

Let $P^{\bullet}$ be the following complex: $0 \rightarrow Hom_{\mathcal{C}} (V,X) \rightarrow Hom_{\mathcal{C}} (V,M \oplus M_0)\rightarrow 0$. Since $X \in add(V)$ and $M_0 \in add(M) \subseteq add(V)$, we can get both $Hom_{\mathcal{C}}(V,X)$ and $Hom_{\mathcal{C}}(V,M \oplus M_0)$ are finitely generated projective $\Lambda$-modules.

I want to ask how to use that $Hom_{\mathcal{C}}(V,X)$ and $Hom_{\mathcal{C}}(V,M \oplus M_0)$ are finitely generated projective $\Lambda$-modules to get the following isomorphism of rings $End_{\Lambda}(T) \cong End_{\mathcal{D}(\Lambda)}(P^{\bullet}) \cong End_{\mathcal{K}(\Lambda)}(P^{\bullet})$?

  • $\begingroup$ Is there a missing assumption, like an exact sequence $0\to X\to M_0\to Y\to 0$? $\endgroup$ – Pierre-Guy Plamondon Aug 31 '16 at 8:49
  • $\begingroup$ @ Pierre-Guy Plamondon 28 $ X \rightarrow M_0 \rightarrow Y$ is $add(M)$-exact sequence in $\mathcal{C}$, but I don't think this is very useful to get I want, maybe I can add it to my question. $\endgroup$ – Xiaosong Peng Aug 31 '16 at 9:24

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