Show that $S^2$ is not diffeomorphic to the Torus.
There are lot of duplications of this question in MSE and I have seen an answer that uses Euler numbers. But let me describe how differ this post from them before marking this as a duplicate. I found this exercise in Guillemin Pollack Differential Topology book under the section of Intersection theory mod 2 of "Transversality and Intersection" chapter. So definitely we have to use some facts or property on intersection mod 2. But honestly I have no clue what to use/ how to start... Can anyone please help me to figure this out?
Let $R>r>0.$ The map $f:\mathbb{S}^1\times \mathbb{S}^1\to \mathbb{R}^3$ given by $$f(e^{i\alpha},e^{i\beta})=((R+r\cos\beta)\cos\alpha,(R+r\cos\beta)\sin\alpha,r\sin\beta)$$ is an embedding and parametrizes the torus. The two circles $f(e^0,e^{i\beta})$ and $f(e^{i\alpha},e^{i\frac{\pi}{2}})$ have intersection exactly the point $(R,0,r),$ so have intersection $1$ mod $2.$ By the other hand, any two curves in $\mathbb{S}^2$ have intersection $0$ mod $2.$ Indeed, any curve in $\mathbb{S}^2$ misses a point by Sard theorem. Since $\mathbb{S}^2-\{p\}\cong \mathbb{R}^2,$ any curve in $\mathbb{S}^2$ is contractible.
This proves that $\mathbb{S}^2$ and the torus are not diffeomorphic.