I am learning about Positselski's co- and contraderived categories. We know that short exact sequences do not generally induce distinguished triangles in the homotopy category but they do in the usual derived category. So I wonder whether a short exact sequence of DG-Modules (or CDG) over a DG-Ring give exact triangles in the co- and/or contraderived category.
1 Answer
Yes, of course. The coderived and contraderived categories (or more specifically, the absolute derived categories) are defined precisely in such a way that short exact sequences of CDG-modules induce distinguished triangles in these triangulated categories.
Let $0\longrightarrow K\overset i\longrightarrow L\overset p\longrightarrow M\longrightarrow0$ be a short exact sequence of CDG-modules. Denote by $T=\operatorname{Tot}(K\to L\to M)$ the total CDG-module of this short exact sequence. By the definition, $T$ is an absolutely acyclic (hence coacyclic and contraacyclic) CDG-module, so it represents a zero object in the absolute derived category (hence in the coderived and contraderived categories).
In the cochain homotopy category of CDG-modules, there is a distinguished triangle $K\overset i\longrightarrow L\overset f\longrightarrow\operatorname{Cone}(i)\overset g\longrightarrow K[1]$. Furthermore, there is a natural closed morphism of CDG-modules $s\colon\operatorname{Cone}(i)\longrightarrow M$ such that $sf=p$. The cone of $s$ is isomorphic, as a CDG-module (up to a shift), to the CDG-module $T$, $$ \operatorname{Cone}(s\colon\operatorname{Cone}(i)\to M)\simeq \operatorname{Tot}(K\to L\to M). $$ Accordingly, in the absolute derived category of CDG-modules, the morphism $s$ gets inverted and becomes an isomorphism.
Therefore, in the absolute derived category of CDG-modules, one can form the triangle $K\overset i\longrightarrow L\overset p\longrightarrow M\overset{gs^{-1}}\longrightarrow K[1]$. This is the desired distinguished triangle attached to the short exact sequence of CDG-modules $0\longrightarrow K\overset i\longrightarrow L\overset p\longrightarrow M\longrightarrow0$. It is indeed a distinguished triangle in the absolute derived category, since the morphism $s$ provides an isomorphism of triangles $(K\overset i\rightarrow L\overset f\rightarrow\operatorname{Cone}(i)\overset g\rightarrow K[1])\longrightarrow(K\overset i\rightarrow L\overset p\rightarrow M\overset{gs^{-1}}\rightarrow K[1])$ in the absolute derived category (while the triangle $K\overset i\longrightarrow L\overset f\longrightarrow\operatorname{Cone}(i)\overset g\longrightarrow K[1]$ is distinguished already in the homotopy category, and consequently in the absolute derived category as well).
As any distinguished triangle in the absolute derived category, the triangle $K\overset i\longrightarrow L\overset p\longrightarrow M\overset{gs^{-1}}\longrightarrow K[1]$ is also distinguished in the coderived and contraderived categories.
-
$\begingroup$ Thank you, that was very enlightening. This makes a lot of the proofs much clearer now. (I apologize for misspelling your name) $\endgroup$– So LetCommented Feb 5 at 20:03