# A an Abelian category, define the derived category as $D(A) = Ch(A)[Qiso^{-1}]$. Why do homotopic maps become equal?

I am defining $D(A)$ by the localization of Ch(A) for $A$ an abelian category at the set of quasi-isomorphisms, and $Q : Ch(A) \to D(A)$ is the localization map.

(So $D(A)$ for this question means: the objects are the same as Ch(A), but we have introduced for each quasi-isomorphism and "edge" between the domain and codomain of that quasi-isomorphism. Morphisms are paths of edges that are subject to certain equivalence relationships, namely: if there are two adjacent edges that are honest morphisms in $Ch(A)$, we can compose them and replace the length 2 piece of that path by their composition. If the fake inverse to a quasi-isomoprhism $s$ is placed next to $s$, we can replace them by the identity morphism.

$Q$ then sends objects to objects, and on morphisms it sends $f : X \to Y$ to the path $f$.)

This construction does not first construct the homotopy category $K(A)$, so this question is not a complete tautology. (It is the relationship between these two constructions of the derived category that I am trying to understand.)

Suppose that $f : X \to X$ is homotopic to the identity. Why is $Q(f) = 1_X$ in the derived category?

I ask because the proof that $Q(h) = Q(g)$ if $h$ and $g$ are homotopic in Gelfand-Manin (page 159, chapter 3 part 4) seems to reduce everything to this case, and I don't see why it is true.

It suffices to show that a nullhomotopic map is sent to zero in the derived category, since the localization functor is additive (I think). But I am at a loss for how to describe chain homotopies in the derived category.

If $X$ is a chain complex, let $X'$ denote the chain complex with $X'_n=X_n\oplus X_n\oplus X_{n-1}$ and $d_n^{X'}:X'_n\to X'_{n-1}$ given by the matrix $$\begin{pmatrix} d_n^X & 0 & (-1)^{n-1} \\ 0 & d_n^X & (-1)^n \\ 0 & 0 & d_{n-1}^X\end{pmatrix}$$
(To motivate this construction, if $A=Ab$, then $X'$ is just the tensor product of $X$ with the chain complex $\mathbb{Z}\stackrel{(1,-1)}\to\mathbb{Z}^2$ in degrees $1$ and $0$, which you can think of geometrically as the chain complex of the simplicial complex $\Delta^1$. You should think of $X'$ as an algebraic version of $X\times \Delta^1$ for a topological space $X$, which of course represents homotopies in topology.)
Let $i_0,i_1:X\to X'$ be the inclusions of the first two summands, and define a map $p:X'\to X$ by $p(x,y,z)=x+y$. (These are the algebraic versions of the inclusions $X\to X\times \Delta^1$ at the two endpoints of $\Delta^1$ and the projection map $X\times\Delta^1\to X$.) It is easy to verify that all of these maps are quasi-isomorphisms, and that $pi_0=pi_1=1_X$. In particular, this implies that $i_0=i_1$ in the derived category, since $p$ is inverted. Note furthermore that a chain-homotopy between two maps $f,g:X\to Y$ is equivalent to a map $h:X'\to Y$ such that $hi_0=f$ and $hi_1=g$ (the chain-homotopy is then given by what $h$ does on the third coordinate of $X'_n$). Since $i_0=i_1$ in the derived category, it follows that $f=g$ in the derived category.