Let \(C\) and \(C'\) be chain complexes, and let \(I\) be the chain complex consisting of \(\mathbb{Z}\) in dimension 1 and \(\mathbb{Z} \times \mathbb{Z}\) 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 \(\mathbb{Z}\)’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 \(h(c) = f(e \otimes c)\).]
Solution
1. \(I \otimes C\) has, in degree \(n\), the group \((I_0 \otimes C_n) \oplus (I_1 \otimes C_{n-1}) = (v_0 \otimes C_n) \oplus (v_1 \otimes C_n) \oplus (e \otimes C_{n-1})\).
\(I_0 = \ZZ v_0 \oplus \ZZ v_1\) and \(I_1 = \ZZ e\), so \(I \otimes C\) in degree \(n\) is \((I_0 \otimes C_n) \oplus (I_1 \otimes C_{n-1})\).
2. The boundary of \(I \otimes C\) is \(\partial(v_i \otimes c) = v_i \otimes \partial c\) and \(\partial(e \otimes c) = (v_1 - v_0) \otimes c - e \otimes \partial c\).
\(\partial(e) = v_1 - v_0\), and the boundary of a tensor product is \(\partial(a \otimes c) = \partial a \otimes c + (-1)^{|a|} a \otimes \partial c\).
3. A chain map \(f: I \otimes C \to C'\) is determined by its values on \(v_0 \otimes c\), \(v_1 \otimes c\), and \(e \otimes c\).
these generate \(I \otimes C\).
4. Define \(f_i(c) = f(v_i \otimes c)\) for \(i = 0, 1\), and \(h(c) = f(e \otimes c)\).
definition.
5. The chain map condition on \(e \otimes c\) gives \(f_1 - f_0 = \partial h + h \partial\).
5.1. \(f(\partial(e \otimes c)) = \partial f(e \otimes c)\).
\(f\) is a chain map.
5.2. \(f(\partial(e \otimes c)) = f((v_1 - v_0) \otimes c - e \otimes \partial c) = f_1(c) - f_0(c) - h(\partial c)\).
2 and 4.
5.3. \(\partial f(e \otimes c) = \partial h(c)\).
4.
5.4. Hence \(f_1(c) - f_0(c) - h(\partial c) = \partial h(c)\), i.e. \(f_1 - f_0 = \partial h + h \partial\).
<2>1–<2>3.
6. The chain map condition on \(v_i \otimes c\) gives that \(f_i\) is a chain map.
\(f(\partial(v_i \otimes c)) = f(v_i \otimes \partial c) = f_i(\partial c)\) equals \(\partial f(v_i \otimes c) = \partial f_i(c)\), so \(f_i \partial = \partial f_i\).
7. Hence \(f\) is exactly a chain homotopy \(h\) between the chain maps \(f_0\) and \(f_1\).
5 and 6.
8. Q.E.D.
7.