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Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : I \to \text{Hom}(C, D)$$

such that the restriction of the map to $0$ is $f$ and the restriction of the map to $1$ is $g$, where $\text{Hom}$ is the hom chain complex. With this definition one can work guided by analogy to the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.

Note that in the topological case it's pretty clear that topological spaces, continuous functions, and homotopies don't form a 2-category for the same reason that closed loops in a space based at a point don't form a group (composition is only associative up to homotopy).

Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : I \to \text{Hom}(C, D)$$

such that the restriction of the map to $0$ is $f$ and the restriction of the map to $1$ is $g$, where $\text{Hom}$ is the hom chain complex. With this definition you can work guided by analogy to the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.

Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : C \otimes I \to D$$

such that the restriction of the map to $C \otimes 0$ is $f$ and the restriction of the map to $C \otimes 1$ is $g$ (it may be necessary to put $I$ on the other side depending on your conventions for tensor products).

  With this definition one can work guided by analogy to the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.

Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : I \to \text{Hom}(C, D)$$

such that the restriction of the map to $0$ is $f$ and the restriction of the map to $1$ is $g$, where $\text{Hom}$ is the hom chain complex. With this definition one can work guided by analogy to the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.

Note that in the topological case it's pretty clear that topological spaces, continuous functions, and homotopies don't form a 2-category for the same reason that closed loops in a space based at a point don't form a group (composition is only associative up to homotopy).

Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : C \otimes I \to D$$

such that the restriction of the map to $C \otimes 0$ is $f$ and the restriction of the map to $C \otimes 1$ is $g$ (it may be necessary to put $I$ on the other side depending on your conventions for tensor products). This is exactly analogous to the topological situation.

With this definition one can work exactly as in the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details.

Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : C \otimes I \to D$$

such that the restriction of the map to $C \otimes 0$ is $f$ and the restriction of the map to $C \otimes 1$ is $g$ (it may be necessary to put $I$ on the other side depending on your conventions for tensor products).

With this definition one can work guided by analogy to the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.