# Homotopy factorization of morphisms of chain complexes

This is a sort of follow up to this MO question.

Let $R$ be a ring (eventually with good properties) and $\mathrm{Chains}(R)$ be the category of chain complexes of $R$-modules (eventually bounded). One has the natural endofunctors $\tau_{\leq n}$ and $\tau_{\geq n}$ associated with the standard t-structure on $\mathrm{Chains}(R)$.

What I'm wondering is the following. Is it true that for every morphism $f\colon A\to B$ of chain complexes there exist morphisms $h:A\to C$ and $g:C\to B$ with $\tau_{\leq -1}(h)\colon \tau_{\leq -1}(A)\to \tau_{\leq -1}(C)$ a weak equivalence (in the standard model structure on chain complexes), $\tau_{\geq 0}(g)\colon \tau_{\geq 0}(C)\to \tau_{\geq 0}(B)$ a weak equivalence, and with $f$ homotopy equivalent to $g\circ h$?

Consider the following diagram in which the rows are cofiber sequences: $$\require{AMScd}\begin{CD} \tau_{\geq0}A @>>> A @>>> \tau_{\leq-1}A @>>>\Sigma\tau_{\geq0}A\\ @VVV @VVV @| @VVV\\ \tau_{\geq0}B @>>> C @>>> \tau_{\leq-1}A @>>>\Sigma\tau_{\geq0}B\\ @| @VVV @VVV @|\\ \tau_{\geq0}B @>>> B @>>> \tau_{\leq-1}B @>>>\Sigma\tau_{\geq0}B \end{CD}$$

Here the vertical maps in the first, third, and fourth columns are the maps induced by $f$, and $C$ can be obtained as the fiber of the rightmost map in the middle row. The second column is then the factorization you seek.

• Hi Eric, thanks! Indeed my candidate for $C$ was $C=\tau_{\leq -1}A\times_{\tau_{\leq-1}B}B$, which I see is the same $C$ you get, but I had been unable to show that $A\to C$ and $C\to B$ became equivalences under $\tau_{\leq -1}$ and $\tau_{\geq 0}$ by the fiber product description. Now I see how looking at the pullback diagram as a part of the big diagram gives that immediately by using $\tau_{\leq -1}\tau_{\geq 0}=0$ and $\tau_{\leq -1}\Sigma\tau_{\geq 0}=0$ and the 3-for-2 property for equivalences. And it also works for an arbitrary t-structure on a triangulated category, right? – domenico fiorenza Jun 3 '14 at 5:32
• Actually, if I'm not mistaken, there is a subtlety that prevents my construction from working for an arbitrary triangulated category with t-structure. Namely, you cannot be sure that the composition $A\to C\to B$ is equal to the original map; all you know is that it is some map that commutes with the top and bottom cofiber sequences (but such maps are not unique). What you need is that everything lifts to an $\infty$-category in which the sequences $\tau_{\geq0}A\to A\to\tau_{\leq-1}A$ are functorial and there is also a functorial nullhomotopy of the composition. – Eric Wofsey Jun 3 '14 at 10:36
• The functoriality of the nullhomotopy makes all the choices of maps in the second column canonical and functorial. – Eric Wofsey Jun 3 '14 at 10:36
• Perfect! a stable $\infty$-category with functorial truncation sequences and nullhomotopies was precisely the context I had in mind! Thanks a lot! – domenico fiorenza Jun 3 '14 at 15:29
• Yeah, I didn't mean to imply that was any sort of harsh restriction. I don't know of any "natural" examples that don't satisfy that. – Eric Wofsey Jun 3 '14 at 15:31

Just to explain Eric's construction in more elementary terms (and to point out that you can get something stronger than in your question). Define $C$ as follows:

$$C_n=\left\{ \begin{array}{ll} B_n,&n>0,\\ A_n&n<0. \end{array} \right.$$ Moreover, let $d_n\colon C_n\rightarrow C_{n-1}$ be the differential of $B$ (resp. $A$) if $n>1$ (resp. $n<0$). We must still define $C_0$ and its surrounding differentials.

We define $C_0$ as the push-out $$\begin{array}{ccc} Z_0(A)&\rightarrow&Z_0(B)\\ \downarrow&&\downarrow\\ A_0&\rightarrow&C_0 \end{array}$$ The upper arrow is the map induced on $0$-cycles by $f$ and the left arrow is the inclusion of $0$-cycles in $0$-chains, which is injective. Therefore the parallel arrow is also injective. Moreover, $A_0/Z_0(A)=C_0/Z_0(B)$.

The differential $d_0\colon C_0\rightarrow C_{-1}=A_{-1}$ is given by applying the universal property of a push-out to $d_0\colon A_0\rightarrow A_{-1}$ and the trivial map $0\colon Z_0(B)\rightarrow A_{-1}$. This shows that $Z_0(C)=Z_0(B)$ and that the images of $d_0\colon C_0\rightarrow C_{-1}=A_{-1}$ and $d_0\colon A_0\rightarrow A_{-1}$ coincide.

The differential $d_1\colon C_1=B_1\rightarrow C_0$ is the composite $B_1\rightarrow Z_0(B)\hookrightarrow C_0$.

We now take $h\colon A\rightarrow C$ to be $f_n\colon A_n\rightarrow B_n$ for $n>0$, the identity for $n<0$, and the bottom map in the push-out square for $n=0$. The map $g\colon C\rightarrow B$ is the identity for $n>0$, $f_n\colon A_n\rightarrow B_n$ for $n<0$, and for $n=0$, the map induced by applying the universal property of a push-out to $f_0\colon A_0\rightarrow B_0$ and $Z_0(B)\hookrightarrow B_0$.

Clearly $f=gh$. Moreover, $\tau_{\geq 0}g$ and $\tau_{\leq -1}h$ are identity maps, not only weak equivalences, by the previous computations. This construction is actually functorial in $f$.

• I have a similar problem here (see my comment to Eric's answer): now I can define $\mathcal E = \left\{ \begin{smallmatrix} A \\ \downarrow \\ B \end{smallmatrix} \mid \begin{array}{ccc} Z_0 A & \overset{Z_0 f}\to & Z_0 B \\ \downarrow && \downarrow \\ A_0 & \underset{f_0}\to & B_0 \end{array} \text{ is a pushout} \right\}$ and $\mathcal M = \left\{ \begin{smallmatrix} A \\ \downarrow \\ B \end{smallmatrix} \mid \begin{array}{ccc} Z_0 A & \overset{Z_0 f}\to & Z_0 B \\ \downarrow && \downarrow \\ A_0 & \underset{\cong}\to & A\amalg_{ Z_0A}Z_0B \end{array} \right\}$. Is it a FS? – Fosco Loregian Jun 4 '14 at 13:24
• @tetrapharmakon I don't think so, too few conditions. – Fernando Muro Jun 4 '14 at 15:49
• Ha! So this doesn't solve the original problem :( (your answer was extremely useful, though) – Fosco Loregian Jun 4 '14 at 16:02
• Oh. I see now that Domenico phrased the original problem in a form weaker than that we need. In fact we were looking for factorization systems on $Ch(R)$, not only for factorization functors (thought as sections of the composition functor $Ch(R)\times Ch(R)\to Ch(R)$)... Can your proof (or Eric's) be adapted to this end? – Fosco Loregian Jun 4 '14 at 16:11
• @tetrapharmakon I just meant that $\mathcal E$ and $\mathcal M$ are not the appropriate classes of morphisms to consider. More conditions are necessary ($A\rightarrow B$ must induce isomorphisms in negative/positive degrees). – Fernando Muro Jun 4 '14 at 17:27