Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
up vote 12 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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.