0

$\newcommand{\del}{\partial}$This is the definition of chain homotopy I am using:

Two chain maps $f,g$ are chain homotopic iff $f-g=\del T + T\del$ for some chain map $T$.

  • For any two chain maps $f-g$ is also a chain map, so $\del(f-g)=(f-g)\del$. But if $f,g$ are chain homotopic, we have additionally that $\del(f-g)=(f-g)\del=0$, and thus that $\del f=\del g = g\del=f\del$.
  • Since $T$ is a chain map, $\del T+T\del =2\del T = 2T\del$ (2 here means iterated addition). In particular, $2T$ is a chain map.

Question: The above conditions are necessary for $f,g$ to be chain homotopic, but are they sufficient? I.e., does: $$\del f=\del g =g\del=f\del \implies \exists\text{ a chain map } \psi \text{ such that }f-g=\del\psi?$$ This sort of seems like it would be "integrating" $f-g$, and since I have no special assumptions, I assume as a result that the answer is no. (E.g. the same way that not all closed forms are exact.)

Also, why is the condition for $f$ and $g$ to be chain homotopic not stated as $f-g=\del\psi$ for some chain map $\psi$, since that's all that we need for the homomorphisms of $f$ and $g$ to coincide on homology? (I am aware that most likely not every chain map $\psi$ can be decomposed as $\psi=T+T$ for another chain map $T$, so that $f-g=\del\psi=\psi\del$ does not imply $ f-g = \del (T+T)= \del T + T\del$.)

Would it be correct to say that $f-g=\del\psi$ for some chain map $\psi$ means that they are "chain homologous" (i.e. $f$ and $g$ induce the same homomorphisms on homology, but the condition is necessary but not sufficient to be chain homotopic)?

Chill2Macht
  • 20,920
  • 2
    First of all, your definition of chain homotopy seems to be wrong: $T$ is not a morphism of chain complexes, it's a map that shifts the degree by $1$, i.e. $T^n\colon C^n \to D^{n-1}$ (in cohomological numeration). –  Dec 06 '16 at 15:21

1 Answers1

2
  1. If $f^\bullet, g^\bullet\colon C^\bullet\to D^\bullet$ are chain maps, then the correct definition is the following.

A chain homotopy between $f^\bullet$ and $g^\bullet$ is a family of morphisms $T^n\colon C^n\to D^{n-1}$ such that for all $n\in \mathbb{Z}$ holds $$f^n - g^n = \partial^{n-1}_D \circ T^n + T^{n+1}\circ \partial^n_C.$$

$T^\bullet$ is not a chain map in any sense. The definition just tells that $\partial^{\bullet-1}_D \circ T^\bullet + T^{\bullet+1}\circ \partial^\bullet_C$ is a chain map, which is equal to $f^\bullet - g^\bullet$.

enter image description here

  1. Chain maps that induce the same maps in (co)homology are not necessarily homotopic. Probably the easiest example is the following: fix some $n=2,3,4,\ldots$ and consider the following complex of abelian groups $$C^\bullet\colon \quad 0 \to \mathbb{Z} \xrightarrow{\times n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$$ It is an exact sequence, its (co)homology is trivial, hence the identity morphism $id$ on $C^\bullet$ induces the same morphisms on (co)homology as the zero morphism $0$. But it is easy to see that there is no chain homotopy between $id$ and $0$:

enter image description here

  • This is a good counterexample and makes a lot more sense to me now. Just to confirm I understand your answer before accepting it, is the condition I gave above for two chain maps to induce the same map in homology correct? I was able to prove it was sufficient but not that it was necessary. – Chill2Macht Dec 07 '16 at 09:09
  • This is sort of how I am interpreting this proofwiki page but im not sure: https://proofwiki.org/wiki/Homotopic_Chain_Maps_Induce_Equal_Maps_on_Homology – Chill2Macht Dec 07 '16 at 09:16
  • also I should clarify that when I said "chain map" (I don't know if this definition is correct) I meant sequences of homomorphisms of chain complexes which commute with the boundary operator of any degree (not necessarily zero) – Chill2Macht Dec 07 '16 at 09:48
  • 1
    @William A chain map is a collection of maps $f^n\colon C^n \to D^n$ that commute with the boundary operators in all degrees, that's correct. However, a chain homotopy is not a chain map, and when you write identities like "∂T + T∂ = 2∂T = 2T∂", it's wrong. –  Dec 07 '16 at 15:16
  • 1
    @William I recommend you to write the involved identities with all the indices to see what they really mean degree-wise. You are asking about "$f-g = \partial \psi$" where $\psi$ is a chain map. But if $\psi$ is really a chain map (= a collection of maps $\psi^n\colon C^n\to D^n$ commuting with the boundary operators), then the composition $\partial^n \circ \psi^n$ shifts the degree by one, it's not a chain map and it makes no sense to compare it to $f^n - g^n$. –  Dec 07 '16 at 15:26