Cup products in cohomologies of connected sum of copies of a specific manifold Cup products in cohomologies of connected sum of copies of $S^3\times S^4$
I want to calculate $H^*\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)$ (that is, connected sum of 5 copies of the same space).
Since 
$$H^*\left(S^3,\mathbb{Z}\right)=\mathbb{Z}<1>\oplus0\oplus0\oplus\mathbb{Z}<x>
=\mathbb{Z}[x]/_{(x^2)}$$ 
and
$$H^*\left(S^4,\mathbb{Z}\right)=\mathbb{Z}<1>\oplus0\oplus0\oplus0\oplus\mathbb{Z}<y>
=\mathbb{Z}[y]/_{(y^2)},$$
Künneth formula gives us $$H^*\left(S^3\times S^4,\mathbb{Z}\right)=H^*\left(S^3,\mathbb{Z}\right)\otimes H^*\left(S^4,\mathbb{Z}\right)=\mathbb{Z}[x,y]/_{(x^2y^2)},$$ 
where for $n=0,3,4,7~$ $H^n\left(S^3\times S^4,\mathbb{Z}\right)=\mathbb{Z}$ and for $n=1,2,5,6~$ $H^n\left(S^3\times S^4,\mathbb{Z}\right)=0.$
Cohomology group of connected sum is sum of cohomology groups of connected summands for every dimension except for $n=0,7.$ Thus,
\begin{align*}
&H^0\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}\\
&H^1\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^2\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^3\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}=\mathbb{Z}^5\\
&H^4\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}=\mathbb{Z}^5\\
&H^5\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^6\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^7\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}
\end{align*}
If all I did above is OK, it remains to determine cup products, which I am horrible at. Since the groups are like that, we only need to describe the map $$H^3\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right) \times H^4\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right) \xrightarrow{\smile} H^7\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right).$$
How can I determine this map? I found something here https://mathoverflow.net/questions/98376/cup-products-of-connected-sum, but I understood nearly nothing.
Thank you!
 A: It would help to know what was confusing about the MO page. If we have a connected manifold $M \# N$, then the cohomologies are
$H^0(M \# N) = \Bbb Z$
$H^k(M \# N) = H^k(M) \oplus H^k(N) \;\;\;\; 0 < k < n$ 
$H^n(M \# N) = \Bbb Z$. 
I would prove this by using the relative long exact sequence $$H^*(M \vee N) = H^*(M \# N, S^{n-1}) \to H^*(M \# N) \to H^*(S^{n-1});$$ you can find the above description by chasing through the sequence.
The cup product with $1 \in H^0(M \# N)$ is the identity. The cup product between $(m, n)$ and $(m', n')$ whose degrees sum to less than $n$ is $(m \smile m', n \smile n')$. If the degrees sum to $n$, then $$(m,n) \smile (m',n') = m \smile m' + n \smile n'.$$ This will follow again from the above long exact sequence.
So for your space one has $H^3 = \langle e_1, \cdots, e_5\rangle$ while $H^4 = \langle f_1, \cdots f_5\rangle$, with the property that $e_i \smile f_j = \delta_{ij}$, ie, it is zero if $i \neq j$ and 1 if $i = j$. 
That is it's just the direct sum of 5 copies of the cup product pairing of $S^3 \times S^4$.
