The setup is as follows: We have the category $C$ of chain complexes over some additive/abelian category and want to pass to the category $K$ of chain complexes modulo homotopy.

So we have an (additive) functor $F\colon C \to T$, where $T$ is some triangulated category such that $F$ behaves like a triangulated functor (even though $C$ is of course not triangulated), i.e. commutes with the shift functor and maps "triangles" $X\xrightarrow{f}Y\to Cone(f)$ to actual triangles in $T$.

Is it then true, that $F$ factors over $K$?

I was able to show that in the case I am mainly interested in (and in some others), when $C$ is the category of right-bounded complexes of projective modules, then $F$ maps contractible spaces to zero and homotopy equivalences to isomorphism. Here what one mainly uses is that if a map is a homotopy equivalence, then its mapping cone is contractible.

I also seem to recall some statement of the form: a functor factors over the homotopy category if and only if it maps homotopy equivalences to isomorphism, but I can't remember the exact context. Basically I would need this statement to hold for chain complexes.


  • 2
    $\begingroup$ The statement that a functor factors through the homotopy category if and only if it maps homotopy equivalences to isomorphisms holds in great generality, in particular for chain complexes. See the answer of Karol Szumiło to this question: mathoverflow.net/questions/72992/… $\endgroup$ Apr 21, 2014 at 16:06
  • $\begingroup$ I only know how to answer this when $T$ is in fact the homotopy category of a stable model category, e.g. if $T$ is a `topological triangulated category' (Schwede has written extensively about such categories). In that case the answer is yes, by the universal property of the functor that takes a model category to its homotopy category. To realize $K$ as the homotopy category of some model structure on $C$ (with weak equivalences the chain homotopy equivalences rather than quasi-isomorphisms), use this: mathoverflow.net/questions/52508. The linked paper addresses the case you mention. $\endgroup$ Apr 21, 2014 at 16:24

1 Answer 1


Yes, that is true.

I will try to give an elementary proof.

First of all the category $K$ is obtained as a quotient category of $C$ in the sense of [MacLane, Categories for the Working Mathematician II.8]. By Proposition 1 in that very section it is enough to show that for any morphism $f \in \operatorname{Mor} C$, which is homotopic to zero, we have $F(f)=0$.

The important step in showing that $F(f)=0$ is that for $f$ homotpic to zero we have a "section" $s: \operatorname{Cone}(f) \rightarrow Y$ in the triangle

$$ X \xrightarrow{f} Y \xrightarrow{\iota} \operatorname{Cone}(f) \xrightarrow{\pi} X[1], $$ s.t. $s \circ \iota= \operatorname{id}_Y$, where $\iota, \pi$ are the canonical maps and $\operatorname{Cone}(f)= X[1] \oplus Y$ and this triangle is mapped by the functor $F$ to a distinguished triangle.

Define the "section" as a map of chain complexes by $s^{i+1}(x^{i+1}, y^i)= y^i + h^{i+1}(x^{i+1})$ , where $h$ is the homotopy transforming $f$ to zero, i.e. $f^i=d_Y^{i-1} h^i+h^{i+1} d_X^i$ for all $i$.

As needed we have $(s \circ \iota)(y^i)=s(0,y^i)=y^i= \operatorname{id}_Y(y^i)$. More importantly $s$ is actually a map of chain complexes in $C$, since:

$$ s^{i+1} d_{\operatorname{Cone}(f)}(x^i,y^{i-1})=s^{i+1}(-d_X(x^i), f(x^i) + d_Y(y^{i-1})\\ =f(x^i) - h^{i+1}(d_X(x^i)) + d_Y(y^{i-1}) =d_Y(h^i(x^i)) +d_Y(y^{i-1}) \\ = d_Y(s^i(x^i, y^{i-1})) $$

Now consider
$$ F(s) \circ F(\iota) \circ F(f)=F(s \circ \iota) \circ F(f)= F(\operatorname{id}_Y) \circ F(f)= F(f). $$ We will show that the left hand side is the zero map, since $F(\iota) \circ F(f)$ is already zero.

For that consider the distinguished triangle in the triangulated category $T$

$$ F(X) \xrightarrow{F(f)} F(Y) \xrightarrow{F(\iota)} F(\operatorname{Cone}(f)) \xrightarrow{F(\pi)} F(X[1]) $$

The composition of two consecutive maps in a distinguished triangle is zero. This is a general fact about triangulated categories see [Gelfand, Manin, Methods of Homological Algebra IV.1.3 Proposition]. They show that for all objects $A \in T$ the functor $\operatorname{Hom}(A,-)$ is cohomoligical, so in particular for a distinguished triangle

$$ A \xrightarrow{a} B \xrightarrow{b} C \xrightarrow{c} A[1] $$ the sequence $$ \operatorname{Hom}(A,A) \xrightarrow{a_*} \operatorname{Hom}(A,B) \xrightarrow{b_*} \operatorname{Hom}(A,C) $$ is exact and so $0=(b_* \circ a_*)(\operatorname{id}_A)= b \circ a \circ \operatorname{id}_A= b \circ a$. This applied to the above distinguished triangle will give us $F(\iota) \circ F(f)=0$, what we needed to show.

In fact the resulting functor $\bar F: K \rightarrow T$ is exact (functor of triangulated categories), if $F$ is not only addiditve, but also commutes with the shift functor. This is exactly the Claim in [Deligne/Verdier, Categories derivees, etat 0 in Lecture Notes in Mathematics 569 aka SGA 4 1/2 p. 262ff 2-4.]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.