Non homeomorphism I want to show that the sphere $S^2$ and the torus $T^2$ are not homeomorphic, using the notion of intersection modulo $2$.
I have to show that any two loops on the sphere $S^2$ have an even number of intersection points ($=0[2]$) while the torus has loops with an odd number of intersection points. Then, conclude that $S^2$ and $T^2$ are not homeomorphic.


*

*Since $S^2$ is simply connected. I do not know yet how to explore the Van Kampen theorem, I'm bad in group theory (free groups, amalgamated product ...) do you have a good reference for this?


I use a result (from Van Kampen) which states that the union of two simply connected, whose intersection is path connected is simply connected: $S^2=U\cap V$ where $U=S^{2}-\{N\}$ is homeomorphic to $\mathbb{R}^2$ (thus simply connected), the same for $V=S^2-\{-N\}$ and $U\cap V$ homeomorphic to $\mathbb{R}^2-\{0\}$ so path connected.
From two disjoint loops (null intersection) and deforming one using a homotopy, so the number of intersection mod.2 remains invariant $=0[2]$.
Is the intersection mod.2: $\sharp(X\cap Y)[2]:=(i,Y)[2]$ homotopy invariant, or only stable in the sense
$$\exists\varepsilon> 0,\ \forall t\in[0,\varepsilon],\ (i_t,Y)[2]=(i,Y)[2]$$
where $i_t$ is a homotopy with $i_0=i$?


*Using the parameterization of the torus of revolution $\varphi(\alpha,\beta)=((R+r\cos\alpha)\cos\beta,(R+r\cos\alpha)\sin\beta,r\sin\alpha)$ with $0<r<R$, we see that the two circles: $c_1=\varphi(\alpha,0)$ and $c_2=\varphi(\frac{\pi}{2},\beta)$ intersect at exactly one point $(R,0,r)$.


Why this prohibits the existence of an homeomorphism between $S^2$ and $T^2$? 
Is the following reasoning correct:
If $ f: T^2\to S^2$ is a continuous bijection, then $f(c_1)$ and $f(c_2)$ are two loops on $S^2$ with a single point of intersection, which is impossible on $S^2$.
Thanks for your help (and sorry for my bad English).
 A: One way to do this is to use differential topology, and to prove $S^2$ and $T^2$ are not diffeomorphic, which is technically a weaker statement (though it does happen to be equivalent in dimension two). I'll make comments on homeomorphism after.
Intersection numbers mod 2 can be defined in the smooth category using transversality and are well-defined up to smooth homotopy. The intersection number mod 2 of two curves is defined to be the number, modulo two, of intersections of perturbations of them which are transverse. This is well-defined, using transversality methods. See Topology from a differentiable viewpoint by Milnor for a very nice exposition.
To show any two curves on $S^2$ have zero intersection mod two, the Van Kampen theorem may be used as you discuss, but alternatively, since we're in the smooth category, one can use Sard's theorem to show a a smooth curve misses a point, and is thus contractible, for $S^2 - \{\mathrm{pt}\} \cong \mathbb{R}^2$. (This can actually be used to give another proof of the contractibility of any curve, even just continuous, using Weierstrass approximation to homotop a continuous curve to a smooth one, though we don't need it).
Thus $S^2$ and $T^2$ are not diffeomorphic.

How do we do this in the continuous category?
Given two continuous images of $S^1$ on your two-manifolds $S^2$ and $T^2$, define their intersection number to be the minimum number of intersections of any pair of curves homotopic to them. (You don't even need mod 2!) Let's say we permit $\infty$ for the moment (though it turns out always to be finite).
On $S^2$ you showed this is always zero, for any pair of curves.
On $T^2$, we have a pair of curves that intersect once, which you describe. We'd like to show they can't intersect zero times.
Method 1: One way to proceed is to use the smooth result, actually: If they could be homotoped to intersect zero times, these images would be compact and disjoint and thus there would exist some positive minimum distance between them. Then we could approximate these by smooth maps which intersect zero times as well as the whole homotopy by a smooth map (i.e. approximate a map from $S^1 \amalg S^1$ and a map from $(S^1 \amalg S^1) \times [0,1]$) and use the smooth result to get a contradiction.
Thus $S^2$ and $T^2$ are not homeomorphic.
Method 2: These minimum intersection numbers of curves on surfaces can also be computed fully by homotopy and covering space methods. (Such methods much more quickly prove $S^2$ is not homeomorphic to $T^2$, without intersection numbers, of course.)
