Product structure in Cohomology Spectral sequence In the Serre Spectral sequence, we know, the cup product structure induces a canonical product in all $E_{r}$ pages which is compatible with respect to the differential.  I am trying to understand this product in $E_2$ page.  $E_2^{p,q}= H^{p}(B,H^{q}(F,R)) \times H^{r}(B,H^{s}(F,R))=E_2^{r,s} \rightarrow H^{p+r}(B,H^{q+s}(F,R))=E_2^{p+r,q+s} $ has the description as follows, if $\phi, \psi$ are two cocycles , the we define the product as $\phi \cup \psi$ and the coefficients are multiplied by the cup product structure in $H^{*}(F,R)$. 
I believe, on the singular level, it means, for any ,p+r-singular simplex $\sigma$,  $ \phi \cup \psi (\sigma)= \phi(\sigma|[v_0,v_1,...,v_p]) \cup \psi(\sigma|[v_p,..,v_{p+r}])$. Is my interpretation correct?  Second question what does product mean exactly? I have seen, in some computation, Hatcher uses the fact product of two generator is again a generator? Why can the product not be some multiple of generator in the codomain? Am I missing something ? Kindly say some few words on the confusion.  Thank you. 
 A: For abelian groups $G$ and $G'$, there is a map $$H^n(X;G)\otimes H^m(X;G')\to 
H^{n+m}(X\times X;G\otimes G')\xrightarrow{\Delta^*}H^{n+m}(X;G\otimes G')$$ for any space $X$. Recall there is a homotopy equivalence of chain complexes $f:C_*(X\times X)\simeq C_*(X)\otimes C_*(X)$ by Eilenberg-Zilber. The first map in the sequence above sends cochains $\phi:C_n(X)\to G$ and $\phi':C_m(X)\to G'$ to the cochain given by the composition $$C_{n+m}(X\times X)\xrightarrow{f_{n+m}}\bigoplus\limits_{i+j=n+m}C_i(X)\otimes C_j(X)\xrightarrow{\text{proj}}C_n(X)\otimes C_m(X)\xrightarrow{\phi\otimes\phi'}G\otimes G'$$Now suppose there is a group homomorphism $\mu:G\otimes G'\to G''$, for example the cup product $$H^i(Y;R)\otimes H^j(Y;R)\to H^{i+j}(Y;R)$$ for a space $Y$ and a ring $R$, then we get a map $$H^{n+m}(X;G\otimes G')\to H^{n+m}(X;G'')$$ via
$$\bigg(\phi:C_{n+m}(X)\to G\otimes G'\bigg)\mapsto\bigg(\phi:C_{n+m}(X)\to G\otimes G'\xrightarrow{\mu}G''\bigg)$$
edit: I should remark that if we set $G=G'=G''=R$ for $R$ a ring and $\mu$ the multiplication in $R$, this is the ordinary cup product.
