A relatively simple question. A book I'm reading states "a complex homotopic to the zero complex is called contractible"... but I don't understand the statement.
I know what it means for chain maps to be homotopic, but not chain complexes themselves. What does it mean for a complex to be homotopic to the zero complex?
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$\begingroup$Well the statement isn't very precise. Indeed, it's maps that can be homotopic, while complexes can be homotopy equivalent. Two complexes $C$, $D$ are homotopy equivalent if there exists chain maps $f : C \to D$ and $g : D \to C$ such that $f \circ g$ is homotopic to $\operatorname{id}_D$ and $g \circ f$ is homotopic to $\operatorname{id}_C$.
Finally, a chain complex $C$ is said to be contractible if it is homotopy equivalent to the zero complex. If you unpack the definition, it means that there exists linear maps $h : C_n \to C_{n+1}$ such that for all $x \in C$, $$x = dh(x) + h(dx).$$
Note that this is not equivalent to having vanishing homology (a condition called acyclicity). For example, the following chain complex is acyclic but not contractible: $$\dots \to 0 \to \mathbb{Z} \xrightarrow{2 \cdot} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$ Indeed if it were contractible, then the exact sequence would split, which isn't the case.
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