Showing the equality between chain map and chain homotopy This is one of the exercise problem in Hatcher.

Let $C$ and $C'$ be chain complexes and let $I$ be the chain complex consisting of $\Bbb Z$ in dimension $1$ and $\Bbb Z\times\Bbb Z$ in dimension $0$, with the boundary map taking a generator $e$ in dimension $1$ to the difference $v_1-v_0$ of generators $v_i$ of the two $\Bbb Z$'s in dimension $0$. Show that a chain map $f:I\otimes C\to C'$ is precisely the same as a chain homotopy between the two chain maps $f_i:C\to C'$, $c\mapsto f(v_i\otimes c)$, $i =0,1$. [The chain homotopy is $h(c) = f(e\otimes c)$]

The fact that $h(c) = f(e\otimes c)$ is a chain homotopy between $f_1$ and $f_2$ is easy to show. The problem is to show $f\equiv h$. Well from definition, $(I\otimes C)_n = (I_0\otimes C_n)\oplus(I_1\otimes C_{n-1})$. Hence the elements in $(I\otimes C)_n$ is a linear combination of $v_0\otimes c,v_1\otimes c,e\otimes c$ where $c\in C_n$ or $c\in C_{n-1}$. Hence the behavior of $f$ is completely determined by $f(v_0\otimes c),f(v_1\otimes c)$ and $f(e\otimes c)$. Since $I_1=\Bbb Z$, I can identify $I_1\otimes C_{n-1}\simeq C_{n-1}$. So on $I_1\otimes C_{n-1}$, $f$ is same as $h$.
Now I need to show $f$ is same as $h$ on $I_0\otimes C_n$. But I wonder if this make senses. To me, the only possible explanation is $f(v_i\otimes c) = 0$ for any $c\in C_n$ and $i=0,1$. Is this true? Did I understand this problem correctly?
Note. To $f$ be a chain map, if $\partial':C'_n\to C'_{n-1}$ and $\partial:(I\otimes C)_n\to(I\otimes C)_{n-1}$ are boundary maps, $\partial'f(v_0\otimes c) = f(v_0\otimes \partial c)$.
 A: It is an abuse of notation to say that a chain map $f:I\otimes C\to C'$ is precisely the same as a chain homotopy between the two chain maps $f_i:C\to C'$, $c\mapsto f(v_i\otimes c)$, $i =0,1$. We should correctly express it as below. But let us first understand $I\otimes C$. As you say, we have $$(I\otimes C)_n = (I_0 \otimes C_n)  \oplus (I_1 \otimes C_{n-1}) .$$
The boundary maps on $I\otimes C$ are expressed via the boundary maps $\partial : C_n \to C_{n-1}$ as
$$\bar \partial : (I\otimes C)_n \to (I\otimes C)_{n-1}, \bar \partial(v_i \otimes c) = v_i \otimes  \partial c, \bar \partial(e \otimes c) = (v_1-v_0) \otimes c - e \otimes \partial c \\= v_1 \otimes c - v_0 \otimes c - e \otimes \partial c.$$
Let us define chain maps
$$j_i : C \to I\otimes C, j_i(c) = v_i \otimes c .$$
These are in fact chain maps because $\bar \partial j_i(c) = \bar \partial(v_i \otimes c) = v_i \otimes \partial c = j_i \partial c$. In particular, each chain map $f : I\otimes C\to C'$ gives us two chain maps $f \circ j_i : C \to C'$.
Given two chain maps $f_i : C \to C'$, we let $\mathcal{CH}(f_0,f_1)$ denote the set of chain homotopies from $f_0$ to $f_1$ and $\mathscr H(f_0,f_1)$ denote the set of chain maps $f : I\otimes C\to C'$ such that $f \circ j_i = f_i$. Then the correct theorem is this:
The map $\phi : \mathscr H(f_0,f_1) \to \mathcal{CH}(f_0,f_1), \phi(f)(c) = f(e \otimes c)$, is a bijection.
You have already verified that in fact $h = \phi(f)$ is a chain homotopy from $f_0$ to $f_1$. Let us describe an inverse for $\phi$. A chain homotopy $F : f_0  \simeq f_1$ is a collection of $F_n : C_n \to C'_{n+1}$ such that $\partial' F_n  + F_{n-1}\partial = (f_1)_n - (f_0)_n$. Define
$$\psi :\mathcal{CH}(f_0,f_1) \to \mathscr H(f_0,f_1), \psi(F)_n(v_i \otimes c) = (f_i)_n(c), \psi(F)_n(e \otimes c) = F_{n-1}(c) .$$

*

*$\partial' \psi(F)_n = \psi(F)_{n-1} \bar \partial$, i.e. $\psi(F)$ is a chain map $I \otimes C \to C'$:
a) For the elements $v_i \otimes c$ with $c \in C_n$ we have $$\partial' \psi(F)_n(v_i \otimes c) = \partial' (f_i)_n(c) = (f_i)_{n-1}\partial c = \psi(F)_{n-1} (v_i \otimes\partial  c)\\ = \psi(F)_{n-1}\bar \partial (v_i \otimes c)$$.
b) For the elements $e \otimes c$ with $c \in C_{n-1}$ we have $$\partial' \psi(F)_n(e \otimes c) = \partial' F_{n-1}(c) = (f_1)_{n-1}(c) - (f_0)_{n-1}(c) - F_{n-2}\partial(c)\\ = \psi(F)_{n-1}(v_1 \otimes c - v_0 \otimes c - e \otimes \partial c) = \psi(F)_{n-1}\bar \partial (e \otimes c) .$$


*$\psi (F) \circ j_i = f_i$: This is immediate from the definitions.


*$\psi \circ \phi = id$, i.e. $\psi (\phi(f)) = f$ for all $f$:
$\phi(f)$ is a chain homotopy from $f_0$ to $f_1$ and $f ,\psi (\phi(f))$ are chain maps $I\otimes C\to C'$. We have $\psi (\phi(f))(v_i \otimes c) = f_i(c)  = (f \circ j_i)(c) = f(v_i \otimes c)$ and $\psi (\phi(f))(e \otimes c) = \phi(f)(c) = f(e \otimes c)$.


*$\phi \circ \psi = id$, i.e. $\phi (\psi(F)) = F$ for all $F$:
$\psi(F)$ is a chain map $I\otimes C\to C'$ and $F, \phi (\psi(F))$ are chain homotopies. We have $\phi (\psi(F))(c) = \psi(F)(e \otimes c) = F(c)$.
