Question about the definition of chain homotopy

Question

I recently learned about the definition of chain homotopy.

1. If $f^\bullet, g^\bullet\colon C^\bullet\to D^\bullet$ are chain maps, then the 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 $$f^n - g^n = \partial^{n-1}_D \circ T^n + T^{n+1}\circ \partial^n_C.$$

My question: What i do not understand is, what exactly the difference or the sum of maps in $$f^n - g^n = \partial^{n-1}_D \circ T^n + T^{n+1}\circ \partial^n_C.$$ are supposed to mean. What does $f^n-g^n$ mean in this diagram? What does the sum on the right hand side of the equation then tell me? Or generally, what does the above equation exactly describe?

Thanks for any help!

Answer 1

Both $f^n$ and $g^n$ are maps $C^n \to D^n$, thus their difference $f^n-g^n$ is defined by $$(f^n-g^n)(x)=f^n(x)-g^n(x)$$ for any $x \in C^n$. Similarly, the right hand side of the equation is defined by $$(\partial_D^{n-1} \circ T^n + T^{n+1} \circ \partial_C^n)(x) =\partial_D^{n-1} (T^n (x)) + T^{n+1} (\partial_C^n (x))$$ for any $x \in C^n$.

The utility of two maps being chain homotopic is that they induce the same map on cohomology. Since $f^{\bullet}$ and $g^{\bullet}$ are chain maps, they induce well defined maps on cohomology $$\tilde{f}^{\bullet}, \tilde{g}^{\bullet} : H^{\bullet}(C) \to H^{\bullet}(D)$$ defined by $$\tilde{f}^{n}([x])=[f^n(x)]$$ for any $[x] \in H^n(C)$, and similarly for $\tilde{g}^{\bullet}$. Now, since we know $f$ and $g$ are homotopic, for any $[x] \in H^n(C)$ and any representative $x$ of the class $[x]$, we have $$[(f^n-g^n)(x)]=[\partial_D^{n-1} (T^n (x))] + [T^{n+1} (\partial_C^n (x))]$$ which simplifies to $$[f^n(x)-g^n(x)]=[\partial_D^{n-1} (T^n(x))],$$ as $\partial_C^n(x)=0$, since $x$ represents a cohomology class. Therefore $$[f^n(x)]=[g^n(x)] +[\partial_D^{n-1}(T^n(x))]$$ and since $[\partial_D^{n-1}(T^n(x))]$ is the trivial cohomology class, we have $$[f^n(x)]=\tilde{f}^n([x])=\tilde{g}^n([x])=[g^n(x)].$$ That is, $\tilde{f}^{\bullet}=\tilde{g}^\bullet$ as maps $ H^n(C) \to H^n(D)$.