Question about induced cap product on homology/cohomology I have a question regarding the definition of the cap product in Hatcher's algebraic topology. He first defines the cap product on the chain/cochain complexes. He then uses the identity:
$$\partial (\sigma \cap \phi)=\pm (\partial \sigma\cap\phi-\sigma\cap\delta\phi)$$
In order to show that the cap product of a cycle and a cocycle is a cycle, the product of a cycle and a coboundary is a boundary and the product of a cocycle and a boundary is a boundary.
He then claims that this implies the the cap product is also induced in the level of homology/cohomology. I am not sure I understand why that is. Usually when we want to induce a map between homologies/cohomologies (separately), we show that the map sends (co)cycles to (co)cycles and (co)boundaries to (co)boundaries. But this map mixes between homology and cohomology, so I'm not sure what exactly needs to be shown in order to see that this map respects the equivalence classes (I tried writing the cap product explicitly in terms of classes to see why, but I am fairly new to this product so I can't really see why it works out).
Could anyone explain why the three facts above imply that the cap product is induced in the level of homology/cohomology?
Thanks in advance.
 A: First of all in order for the cap product to make sense on (co)homology, we need that the cap product of a cycle and a cocycle is a cycle. Why is this? Let the cycle $z$ represent $[z]$ and the cocycle $\phi$ represent $[\phi]$. The natural way to extend the the cap product to (co)homology is to define
$$[z]\smallfrown [\phi]:=[z\smallfrown \phi]$$ Hence, $z\smallfrown \phi$ better be a cycle, otherwise it does not represent a homology class.
We however need something more if we want this definition to be well-defined. Suppose that $[z]=[z']$, we want their cap product with $[\phi]$ to give the same homology class. In other words, we need $$[z\smallfrown\phi]=[z'\smallfrown\phi]$$
If $[z]=[z']$, then the cycles $z$ and $z'$ differ by a boundary $b$, i.e. $z=z'+b$. Now $$[z\smallfrown\phi]=[(z'+b)\smallfrown\phi]=[z'\smallfrown\phi+b\smallfrown\phi]$$ and if $b\smallfrown\phi$ is a boundary, then this will equal $[z'\smallfrown\phi]$.
Hopefully you know also see why we need that the cap product of a cycle and a coboundary gives a boundary.
