6
$\begingroup$

Consider complexes $(A,d_1), (A',d_1)$, $(C,d_2), (C',d_2)$ and morphisms $f_1,f_2: (A,d_1)\to (A',d_1)$ and $g_1,g_2: (C,d_2)\to (C',d_2)$ of degrees $0$. Consider the functor $(-\otimes-)$, then $$A\otimes C = \bigoplus_{m,n} A^m\otimes C^n, \text{along with differentials}$$ $$ \partial_1 = d_1\otimes C: A^n\otimes C^m\to A^{n+1}\otimes C^m, $$ $$ \partial_2 = (-1)^n A\otimes d_2: A^n\otimes C^m\to A^n\otimes C^{m+1} $$ forms a double complex.

Given homotopies $s:f_1\cong f_2$ and $t:g_1\cong g_2$, Cartan Eilenberg claims that $(s\otimes C, (-1)^nA\otimes t)$ yields a homotopy between $f_1\otimes g_1$ and $f_2\otimes g_2$ which I failed to see. As a first step, we need to check that $$ (s\otimes C)\partial_1+\partial_1(s\otimes C)+ (-1)^n(A\otimes t)\partial_2+ (-1)^n\partial_2(A\otimes t) = f_1\otimes g_1-f_2\otimes g_2 $$ but this is not true, as the left hand side simplifies to \begin{align*} & (s\otimes C)\partial_1+\partial_1(s\otimes C)+ (-1)^n(A\otimes t)\partial_2+ (-1)^n\partial_2(A\otimes t)\\ = & (sd_1+d_1s)\otimes C+ (-1)^{2n} A\otimes (td_2+d_2t)\\ = &(f_1-f_2)\otimes C+ A\otimes (g_1-g_2) \end{align*} which is not equal to the right hand side.

Could you please help me point out what went wrong? Thank you. The relevant material is page 63 last but one paragraph of Cartan Eilenberg, see attached.


Picture is here

$\endgroup$
2
+100
$\begingroup$

The definition in Cartan-Eilenberg is wrong.

For, $t_i = T(A_1,\ldots, s_i,\ldots,A_r)$ defines a map $T(A_1, \ldots, A_i,\ldots A_r) \to T(A_1, \ldots, A'_i,\ldots A_r)$. But for a homotopy you need a map $T(A_1, \ldots, A_r) \to T(A'_1, \ldots, A'_r)$.

The correct definition of the homotopy in your case is $u=:(s \otimes g_1, f_2 \otimes t): A \otimes C \to A' \otimes C'$. Explicitly $$u(a\otimes c) = s(a) \otimes g_1(c) + (-1)^{\text{deg}(a)}f_2(a) \otimes t(c).$$ Cf. MacLane, Homology, Chap. V, Prop. 9.1 (page 164 in my edition).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.