direct sums in homotopy category

Question

Let $\mathcal A$ be an abelian category satisfying $AB3$(there exist arbitrary direct sums). Then the homotopy category $K(\mathcal A)$ also have direct sums. Here $K(\mathcal A)$ is the category of chain complexes over chain homotopy equivalence classes of maps.

Given a set of complexes $\{A_i\}_{i\in I}$ and their direct sum $\bigoplus_{i\in I} A_i $in complex category $C(\mathcal A)$, I want to prove that $\bigoplus_{i\in I}A_i$ is their direct sum in $K(\mathcal A)$. Suppose we are given $f_i:A_i\rightarrow M$ for $i\in I$, I want to prove that there exists a unique map $f $ in $K(\mathcal A)$ such that $f u_i= f_i$ where $u_i$ is the homotopy equivalence classes of the inclusion map of $A_i$ in $\bigoplus_{i\in I} A_i$.

The difficulty is to prove the uniqueness. Some hints will be appreciated.

Answer 1

You might also interested in the following slightly more conceptual explanation (I don't mean to suggest that this should replace a proof by explicit computation in the beginning, though):

Setup: Given chain complexes $X,Y$ over ${\mathcal A}$, there's the homomorphism complex $\text{Hom}^{\bullet}_{\mathcal A}(X,Y)$ defined by having $\text{Hom}^k_{\mathcal A}(X,Y) := \prod_{i\in{\mathbb Z}} \text{Hom}_{\mathcal A}(X^i,Y^{i+k})$ and differential $\delta(f^{\bullet}) := [\delta,f^{\bullet}] := \delta_Y f_\bullet - (-1)^{k} f_\bullet\delta_X$. Note that this is a chain complex over the category of abelian groups, not over ${\mathcal A}$. Now you can check that the cycles in degree $0$ are given by $\text{Z}^0\text{Hom}^{\bullet}_{\mathcal A}(X,Y) \cong \textbf{Ch}_{\mathcal A}(X,Y)$, and that two of them are homologous if and only if they are homotopic. Hence, for the $0$-th homology you get $\text{H}^0\text{Hom}^{\bullet}_{\mathcal A}(X,Y)\cong\textbf{K}_{\mathcal A}(X,Y)$.

Argument: Now if $(X_i)_i$ is a family of chain complexes with component-wise sum $\bigoplus_i X_i$, you have $\text{Hom}^{\bullet}_{\mathcal A}\left({\bigoplus}_i X_i, Y\right)\cong\prod_i\text{Hom}_{\mathcal A}^{\bullet}(X_i,Y)$ by the universal property of the direct sum, and hence $$\textbf{K}_{\mathcal A}\left({\bigoplus}_i X_i, Y\right)\cong \text{H}^0\text{Hom}^{\bullet}_{\mathcal A}\left({\bigoplus}_i X_i, Y\right)\cong\text{H}^0{\prod}_i \text{Hom}^{\bullet}_{\mathcal A}(X_i, Y)\\\quad\quad\quad\cong {\prod}_i \text{H}^0\text{Hom}^{\bullet}_{\mathcal A}(X_i, Y)\cong{\prod}_i \textbf{K}_{\mathcal A}(X_i, Y)$$ since homology commutes with products in the category of abelian groups.

Answer 2

I think I've solved this problem.

$$\begin{array} A\cdots\longrightarrow & A_i^{n-1}\stackrel{d_i^{n-1}}\longrightarrow &A_i^n \stackrel{d_i^n}{\longrightarrow} & A_i^{n+1} \longrightarrow \cdots\\ &\downarrow{u_i^{n-1}} &\downarrow{u_i^n} &\downarrow{u_i^{n+1}} \\ \cdots\longrightarrow & A^{n-1}\stackrel{d_A^{n-1}}\longrightarrow &A^n \stackrel{d_A^n}{\longrightarrow} & A^{n+1} \longrightarrow \cdots \\&\downarrow{f^{n-1}} &\downarrow{f^n} &\downarrow{f^{n+1}} \\\cdots\longrightarrow & M^{n-1}\stackrel{d^{n-1}}\longrightarrow &M^n \stackrel{d^n}{\longrightarrow} & M^{n+1} \longrightarrow \cdots\end{array}$$ It suffices to show that if $fu_i\sim 0$ ($\sim$ stands for homotopy equivalence) for all $i\in I$, then $f\sim 0$. But when you draw maps $s_i^n:A_i^n\rightarrow M_{n-1}$ for all $i\in I$ and $n\in \mathbb Z$, there are induced maps $s:A^{n}\rightarrow M_{n-1}$ and these maps are what we want in the homotopy equivalence relation.