If $M$ is diffeomorphic to $N$, then $\mathbf{H}_{DR}^p(M) = \mathbf{H}_{DR}^p(N)$.

If $M$ is diffeomorphic to $N$, then $\mathbf{H}_{DR}^p(M) =
\mathbf{H}_{DR}^p(N)$.

I thought I got this, but no.....
Given $\mathbf{H}_{DR}^2(S^3)$ is trivial but $\neq
\mathbf{H}_{DR}^2(T^3)$ is not, how can I show $S^3$ and $T^3$ are not
diffeomorphic?
I am also wondering about the general case, that if $M$ is diffeomorphic
to $N$, then $\mathbf{H}^p_{DR}(M)$ is isomorphic to
$\mathbf{H}^p_{DR}(N)$.
Where $\mathbf{H}^p_{DR}(N)$ is the DeRham Cohomology group.
Thank you...