Given two morphisms between chain complexes $f\_\bullet,g\_\bullet\colon\,C\_\bullet\longrightarrow D\_\bullet$, a chain homotopy between them is a sequence of maps $\psi\_n\colon\,C\_n\longrightarrow D\_{n+1}$ such that $f\_n-g\_n= \partial\_D \psi\_n+\psi\_{n-1}\partial\_C$. I can motivate this definition only when the chain complexes are associated to some topological space. For example if $C\_\bullet$ and $D\_\bullet$ are simplicial chain complexes, then for a simplex $\sigma$, a homotopy between $f(\sigma)$ and $g(\sigma)$ is something like $\psi(\sigma)\approx\sigma\times [0,1]$, whose boundary is $f(\sigma)-g(\sigma)-\psi(\partial\sigma)$.  

But chain homotopy also features in contexts where there is no topological space lurking in the background. Examples are chain homotopy of complexes of graphs (*e.g.* Conant-Schneiderman-Teichner) or chain homotopy of complexes in Khovanov homology. In such contexts, the motivation outlined in the previous paragraph makes no sense, with "the boundary of a cylinder being the top, bottom, and sides", because there's no such thing as a "cylinder". Thus, surely, the "cylinder motivation" isn't the most fundamental reason that chain homotopy is "the right" relation to study on chain complexes. It's an embarassing question, but:



> 
> What is the fundamental (algebraic) reason that chain homotopy is relevant when studying chain complexes?
> 



Why chain homotopy when there is no topology in the background?
Here's one way to look at it: There is a chain complex $Hom(C,D)$ in which the $n$th chain group is the product over $k$ of $Hom(C\_k,D\_{n+k})$ and the boundary is given by $\partial (f(c))=(\partial f)(c)+(-1)^{|f|}f(\partial c)$. A chain map is a $0$-cycle, and two of them are chain homotopic if they differ by a boundary.


EDIT: And then a chain map $B\to Hom(C,D)$ corresponds precisely to a chain map $B\otimes C\to D$, where $B\otimes C$ is defined using the usual convention $\partial (b\otimes c)=(\partial b)\otimes c + (-1)^{|b|}b\otimes \partial c$


