Let $C$ and $C'$ be two chain complexes and $C\otimes C'$ their tensor product. Then define the map

\begin{align} \Theta : H_p(C)\otimes H_q(C')&\to H_{p+q}(C\otimes C')\\ {[z_p]}\otimes {[z_q']}&\mapsto {[z_p\otimes z_q']} \end{align}

What does the following mean:

$\Theta$ is induced by the inclusion map $Z_p\otimes Z_q' \to (C\otimes C')_{p+q}$

And why does this imply that it is natural with respect to chain maps? What does this even mean? (Does it mean that $H_p(-)\otimes H_q(-)\implies H_{p+q}(-\otimes -)$ is a natural transformation between the functors $\textbf{Ch}\times\textbf{Ch}\to \textbf{Ab}$?)


1 Answer 1


Recall the definition of the tensor product of two chain-complexes $(C, \partial_C)$ and $(D, \partial_D)$, namely $$(C \otimes D)_n = \bigoplus_{i = 0} ^n C_i \otimes D_{n-i}$$

so in particular $C_p \otimes D_q \subset (C\otimes D)_{p+q}$. The boundary map $\partial_{C \otimes D}$ is defined on an element $c_p\otimes d_q$ by

$$\begin{align} \partial_{C\otimes D}(c_p \otimes d_q) &= \partial_C c_p\otimes d_q+(−1)^{p}c_p \otimes \partial_D d_q \\&\in (C_{p-1} \otimes D_q)\oplus (C_p \otimes D_{q-1}) \\ &\subset (C\otimes D)_{p+q-1} \end{align}$$

(The boundary map is then extended by linearity to all of $C\otimes D$ because it has a basis consisting of homogenous elements.)

A priori if we restrict this inclusion map to cycles then we get a map $Z(C)_p \otimes Z(D)_q \to (C\otimes D)_{p+q}$, but if $\partial_C(c_p) = 0 = \partial_D(d_q)$ if follows from definition that $\partial_{C\otimes D}(c_p\otimes d_q) = 0$, so $Z(C)_p\otimes Z(D)_q$ is actually in $Z(C\otimes D)_{p+q}$. You should verify for yourself why it follows that $\Theta$ induces a well-defined function on homology groups.

By "natural with respect to chain maps" I think they mean the following: Suppose $\phi_C\colon C\to C'$ and $\phi_D \colon D \to D'$ are chain maps, and so $\phi_C \otimes \phi_D\colon C\otimes D \to C'\otimes D'$ is also a chain map. Then the following diagram commutes:

$$\require{AMScd} \begin{CD} H_p(C)\otimes H_q(D) @>{\Theta}>> H_{p+q}(C\otimes D)\\ @V{[\phi_C]\otimes [\phi_D]}VV @V[\phi_C \otimes \phi_D]VV \\ H_p(C')\otimes H_q(D') @>{\Theta}>> H_{p+q}(C'\otimes D')\\ \end{CD}$$

I haven't written out the details but it looks like verifying commutativity is purely formal.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .