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

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 $$\textbf{Ch}\times\textbf{Ch}\to \textbf{Ab}$$?)

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.