An approach to chain homotopies, alternative to the usual boundary relation, uses the monoidal (closed) structure of $\mathsf{Ch}_\bullet(R\mathsf{Mod})$ with $R$ a commutative ring. In particular, a chain homotopy is an arrow $I\otimes X \rightarrow Y$, where $X,Y$ are chain complexes and $I$ is the interal complex (not the unit object).
Now by the definition of the tensor product in $\mathsf{Ch}_\bullet$ $$(I\otimes X)_n=(I_0\otimes X_n)\oplus(I_1 \otimes X_{n-1})$$ where $I_1,I_0$ are, respectively, free $R$-modules on one and two generators. We can hence further decompose to obtain $$(I\otimes X)_n\cong(I^0_0\otimes X_n)\oplus(I_0 ^1\otimes X_n)\oplus(I_1 \otimes X_{n-1})$$ where $I_0^i$ is the free $R$-module on the genrator $[i]$. Since free $R$-modules on one generator are all isomorphic to the unit object $R$, we can make the identification $$(I\otimes X)_n\cong X_n\oplus X_n\oplus X_{n-1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
Some computations show that the differential given by $$\partial ^\prime(x,y,z)=(\partial x-z,\partial y+z, -\partial z)$$ is nilpotent, hence the "graded thing" we identify with $I\otimes X$ is itself a chain complex.
- How can use (1) to better visualize (the grades of) $I\otimes X$?
- How can I intuitively interpret the formula for the differential?
- How can I see these things geometrically?
- is this identification natural?
For instance, I would have thought about the direct sum $X_n\oplus X_n\oplus X_{n-1}$ as two endpoints and the "line" between them, but the dimensions are the other way around.