# Cup Product Structure on $H^*(M_g)$ from cup product on $S^1\times S^1$

Assuming as known the cup product structure on the torus $$S^1\times S^1$$, compute the cup product structure in $$H^∗(M_g)$$ for $$M_g$$ the closed orientable surface of genus $$g$$ by using the quotient map from $$M_g$$ to a wedge sum of $$g$$ tori.

I know to use the induced map on cohomologies here, but I was wondering if anyone could help me understand the first part of this proof that should make things go much faster:

Since $$M_g$$ is connected, we know that $$H^0(M_g)\cong \mathbb{Z}$$ and that the generator 1 of this group is the identity in the cup product structure. Also, we know that $$H_2(M_g)\cong \mathbb{Z}$$, so $$H^2(M_g)\cong \mathbb{Z}$$ by the universal coefficient theorem for cohomology. Since $$H^n(M_g) = 0$$ for $$n > 2$$, we know that $$\theta \in H^2(M_g), \theta \smile 1 = \theta$$, while its cup product with an element in any other dimension must be $$0$$. So it remains to compute the cup product of elements in $$H^1(M_g)\cong \mathbb{Z}^{2g}$$.

So, I don't really see how the universal coefficient theorem is being used here, and the argument that $$H^n(M_g) = 0$$ for $$n > 2$$ $$\implies \theta\smile 1 = \theta$$ is not clear either. How does this give us the cup product structure for every $$n>2$$?

The cup product makes $$H^*(M_g)$$ into a ring -- in particular, we have $$\alpha \smile 1 = \alpha = 1 \smile \alpha$$ for all $$\alpha$$, always. $$1$$ is just the symbol we use for the multiplicative unit of a ring.
It is a little misleading to say "the generator $$1$$ of this group is the identity in the cup product structure". $$\mathbb{Z}$$ does not have a unique generator as a group; $$1$$ and $$-1$$ both generate $$\mathbb{Z}$$! However, $$H^0(M_g)$$ is always a subring of $$H^*(M_g)$$, and $$\mathbb{Z}$$ has a unique identity as a ring. This ring identity is also necessarily a generator of $$\mathbb{Z}$$ as a group.
Since $$M_g$$ is connected, we know that $$H^0(M_g) \cong \mathbb{Z}$$. Then this group has a generator $$1$$ which is the identity in the cup product structure.
If $$\beta$$ is an element of $$H^k(M_g)$$ for some $$k > 0$$, we have that $$\theta \smile \beta \in H^{k+2}(M_g)$$. Since $$k$$ is positive, $$H^{k+2}(M_g) = 0$$! So, $$\theta \smile \beta = 0$$ for all elements $$\beta$$ in positive degree.