How to prove a map is well-defined. How do I show that $\times_{alg}\colon H_pC_*\otimes H_qD_*\to H_{p+q}(C_*\otimes D_*)$ defined by $\times_{alg}([z]\otimes[w])=[z\otimes w]$ is well-defined?
I say that we suppose $[z]\in H_pC_*=\ker\partial/\text{im}\partial$ and $[w]\in H_pD_*=\ker\partial/\text{im}\partial$ where $\partial$ stands for the differential map.   Furthermore, let $\hat z\in [z]$ and $\hat w\in[w]$. Then there exists $a\in \text{im}\partial$ and $b\in \text{im}\partial $ such that $\hat z=z+a$ and $\hat w=w+b$. This means that $\hat z\otimes \hat w=(z+a)\otimes (w+b)=z\otimes w+z\otimes b+a\otimes w+a\otimes b$. We claim that $z\otimes b+a\otimes w+a\otimes b\in \text{im} d$ where $d$ stands for the differential map.
Is this right? Also, how do I proceed?
 A: I'll denote $\partial_C\colon C_\bullet\to C_{\bullet-1}$, $\partial_D\colon D_\bullet \to D_{\bullet-1}$, and $d\colon (C_\bullet\otimes D_\bullet)_\bullet \to (C_\bullet\otimes D_\bullet)_{\bullet-1}$ the differentials of the respective chain complexes.
You have the right start for proving well-definedness. To finish your argument, you should prove the following lemma as an exercise:

Lemma. Suppose $z\in Z_pC_\bullet$ and $w\in Z_qD_\bullet$. If $z\in B_pC_\bullet$ or $w\in B_qD_\bullet$ then $z\otimes w \in B_{p+q}(C_\bullet\otimes D_\bullet)$. In English, the tensor product of a cycle with a boundary is a boundary.

Hint: Supposing $a = \partial_C(c)$ for $c\in C_{p+1}$, and $w\in Z_qD_\bullet$, can you guess a class $f\in (C_\bullet\otimes D_\bullet)_{p+1}$ such that $d(f) = \partial_C(c)\otimes w$? Further hint: use the definition of $d$ and the fact that $\partial_D (w) = 0$. Then the argument for showing $z\otimes b$ is a boundary for $z\in Z_pC_\bullet$ and $b\in B_qD_\bullet$ is almost the same, but with a sign to take care of.
Now with this lemma you can see that your $a\otimes w$, $z\otimes b$ and $a\otimes b$ are all boundaries.
