Is this an equivalent definition for chain homotopy?

Question

$\newcommand{\del}{\partial}$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)?

Answer 1

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 $f^\bullet - g^\bullet$.

2. 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 $0$: