when does a "triangulated" functor factor over the homotopy category? 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.
Thanks
 A: 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.]
