Show, using the Gauss-Bonnet theorem, that $\mathbb{S}^2$ and $\mathbb{T}^2$ are not diffeomorphic.
I find this exercise a bit strange, since by $\chi(\mathbb{S}^2)\neq \chi(\mathbb{T}^2)$ we obviously have that $\mathbb{S}^2$ and $\mathbb{T}^2$ are not homeomorphic and especially since any diffeomorphism would be a homeomorphism, not diffeomorphic. However, as the exercise suggests, I end up with the result, that the Gaussian curvature of $\mathbb{S}^2$ and $\mathbb{T}^2$ are different. But how can I conclude that they are not diffeomorphic? I mean if I use the Theorema Egregium, we have that there cannot be an isometry, however not every diffeomorphism is an isometry.
If $F : S^2\rightarrow (T^2,g)$ is diffeomorphism, then we have $F^\ast g$ is a metric on $S^2$
Note that $\int_{T^2 } K(g)\ dVol_g =2\pi \chi (T^2)$ Here first term is invariant under isometry So $$ \int_{T^2 } K(g)\ dVol_g = \int_{S^2 } K(F^\ast g) \ dVol_{F^\ast g} =2\pi \chi (S^2)$$ Hence we have a contradiction