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I'm taking a course in algebraic topology, which includes an introduction to (simplicial) homology, and I'm looking for a bit of intuition regarding chain homotopies.

The definitions I'm using are:

Let $f,g : X \to Y$ be continuous functions between topological spaces. A homotopy from $f$ to $g$ is a continuous map $H : X \times [0,1] \to Y$ such that $H(\, \cdot\, , 0) = f$ and $H(\, \cdot\, , 1) = g$.

Let $f_{\bullet}, g_{\bullet} : A_{\bullet} \to B_{\bullet}$ be chain maps between chain complexes $(A, d_A)$ and $(B,d_B)$. A chain homotopy from $f$ to $g$ is a sequence of maps $h_n : A_n \to B_{n+1}$ such that $f_n-g_n=d_Bh_n+h_{n-1}d_A$.

I'm aware of the properties of a chain homotopy and how they are similar to those of a homotopy, but I still find the definition quite opaque and the notion quite hard to picture $-$ it would help me a lot if I could think of a chain homotopy in a similar way to how I think of a homotopy.

Or, to make my question a bit less vague, I'd like to know:

  • What is the rationale behind the definition of a chain homotopy?
  • Is there a fundamental similarity between a chain homotopy and a homotopy, beyond their further consequences?

(General waffle would also be appreciated; I'd really like to develop a good understanding.)

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up vote 14 down vote accepted

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.

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For $A \otimes I$ I get a sign issue, but it works with $I \otimes A$. – Damien L Jul 21 '13 at 15:14

The general idea is:

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

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