$S^3$ and $T^3$ are not diffeomorphic.

Question

Let $f : T^2 \to S^3$ be the smooth map of a 2-torus into $S^3$, therefore $$\int_{T^2}f^*\omega = 0.$$ There is a closed $2$-form $\beta$ on $T^3 = S^1 \times S^1 \times S^1$ and a map $g : T^2 \to T^3$ such that $$\int_{T^2}g^*\beta \neq 0.$$ Use the given information to show $S^3$ and $T^3$ are not diffeomorphic.

So I got that $H^2(S^3)$ is a zero class hence all closed forms are exact. But I couldn't see $H^2(T^3)$ is nontrivial.

Also, it sounds familiar but why trivial and nontrivial cohomology are not diffeomorphic..?

Answer 1

I will write the very detailed explanation for your second question (it is not proof however, for most mathematics students this is the 'obvious' level problem). For the first part, please refer to the comment.

If smooth manifolds $M,N$ is diffeomorphic, then the graded algebra $H^*_{dR}(M)=H^*_{dR}(N)$

[proof] Let $f:M\to N$ be a diffeomorphism and $g:N\to M$ its inverse, then since $f\circ g=id_N$ and $g\circ f=id_M$, we find $id=g^*\circ f^*:H^*(N)\to H^*(N)$ and $id=f^*\circ g^*:H^*(M)\to H^*(M)$.