Can we think of a chain homotopy as a homotopy?

Solution 1:

If $I$ is a chain complex representing an interval, with $I_0 = \mathbb{Z}^2$ and $I_1 = \mathbb{Z}$, with $\partial(x) = (x,-x)$, then a chain homotopy between two maps $f,g : A \to B$ is the same as a map $H : A \otimes I \to B$, where $H(a \otimes (1,0)) = f(a)$ and $H(a \otimes (0,1)) = g(a)$. This explains the "shift" up a dimension in the usual definition you'd see of chain homotopy, since your $h_n : A_n \to B_{n+1}$ corresponds to my $H : A_n \otimes I_1 \to B_{n+1}$.

In general, one kind of homotopy in a model category involves what are called cylinder objects. These are functorial factorizations of the fold map $A \coprod A \to A$ through an object $A'$ that is weakly equivalent (for spaces, an isomorphism on homotopy groups -- for chain complexes, a homology isomorphism) to $A$; the inclusion of $A \coprod A \to A'$ should also be particularly well-behaved (a cofibration). Effectively, you're guaranteeing that two copies of $A$ can play nicely in $A'$, and that $A'$ is a "thickened up" version of $A$ rather than something pathological.

Then a homotopy between two morphisms $f,g : A \to B$ is a map $H : A' \to B$ where the composition $A \coprod A \to A' \to B$ is $f \coprod g$. You'll see this pattern again and again.

Solution 2:

The general idea is:

Homotopic continuous maps between topological spaces induce chain homotopic chain maps between the associated simplicial/singular/whatever chain complexes.