pure sub-complexes of exact subcomplexes

Question

In https://www.google.com/#q=tensor+product+of+complexes%2Benochs a new tensor product of complexes is defined which characterizes flatness in the category of complexes of $R$-modules. That is, a complex $F$ is flat if and only if $F\otimes -$ is exact when $\otimes$ denotes the new tensor product of complexes. Also a sequence $$0\to A\to B \to C \to 0$$ is called pure if $$0\to F\otimes A \to F \otimes B \to F\otimes C \to 0$$ is exact for each complex $F$. It is equaivalent to say that $$0\to Hom(P, A)\to Hom(P, B)\to Hom(P,C)\to 0$$ is exact for every finitely presented complex $P$.

Question: Let $\varepsilon$ be the class of all exact complexes of $R$-modules. IS $\varepsilon$ is closed under pure sub-complexes? (That is pure subcomplexes of exact complexes are exact).

Answer 1

Yes:

Let $0\rightarrow A\rightarrow B \rightarrow C\rightarrow0$ be a pure exact sequence of complexes with $H_*(B)=0$. Consider the complex

$$S^n = \cdots \rightarrow 0 \rightarrow R \rightarrow 0 \rightarrow\cdots,$$

concentrated in degree $n$, and the complex

$$D^{n+1} = \cdots \rightarrow 0 \rightarrow R \stackrel{1}\rightarrow R \rightarrow 0 \rightarrow\cdots$$

concentrated in degrees $n$ and $n+1$ (I'm thinking of differentials going downwards). Let $i_n\colon S^n\rightarrow D^{n+1}$ be the identity in degree $n$. For any complex $X$, $\hom(S^n,X)=Z_n(X)$ are the $n$-cycles, and $\hom(D^{n+1},X)=X_{n+1}$. Moreover, $\hom(i_n,X)$ is the morphism $X_{n+1}\rightarrow Z_n(X)$ induced by the differential. Hence $H_n(X)=0$ if and only if $\hom(i_n,X)$ is an epimorphism.

The complexes $S^n$ and $D^{n+1}$ are finitely presented, hence the rows in the following diagram are exact

$$\begin{array}{ccccccccc} 0&\rightarrow& \hom(D,A)&\rightarrow& \hom(D,B)& \rightarrow& \hom(D,C)&\rightarrow&0\\ &&\downarrow&&\downarrow&&\downarrow&&\\ 0&\rightarrow& \hom(S,A)&\rightarrow& \hom(S,B)& \rightarrow& \hom(S,C)&\rightarrow&0 \end{array}$$ Here the vertical arrows are induced by $i_n$. We are assuming that $\hom(i_n,B)$ is an epimophism for all $n\in\mathbb Z$, therefore so is $\hom(i_n,C)$ by the snake lemma, i.e. $H_*(C)=0$. Now, by the long exact sequence for homology $H_*(A)=0$, hence we are done :-)