In an exercise from an old exam, I found myself confronted with $M=\{(\sqrt{x^2+y^2}-2)^2+z^2=1\}$, $U=M\cap\{x\neq0\vee y>0\}$, $V=M\cap\{x\neq0\vee y<0\}$ and $U\cap V$, all subsets of $\mathbb{R}^3$. It was easy to prove $M$ is a regular submanifold of $\mathbb{R}^3$. Then I was asked to find the De Rham Cohomology of $M,U,V,U\cap V$. By intuition, I came up with: $$\omega_1=\frac{-z\mathrm{d}x+(x-2)\mathrm{d}z}{(x-2)^2+z^2},$$ $$\omega_2=\frac{-z\mathrm{d}x+(x+2)\mathrm{d}z}{(x+2)^2+z^2},$$ basically two angular forms around two points inside the hollow of this torus (for $M$ is a torus) extended by translational invariance in the $y$ direction. That is, to be accurate, I would have to add a $+0\mathrm{d}y$ to those to make them 1-forms over the torus. These are defined on the whole of $M$. Obviously they are closed, but not exact, seen as they come from such a form. I then tried to compute $\omega_1-\omega_2$, and found something which is exact, and has this function as a primitive. I let Wolfram do the computations for me, since they are purely algebraic manipulations. Seen as I do not see any other possible source of cohomology (well, save for other similarly found forms, but I have a hunch these should all be cohomologous), I concluded $U,V,U\cap V$ all have cohomology $\mathbb{R}$. This is contradicted by two things: one, intuition, two, a rather naïve application of the Künneth formula. Intuition tells me $U\cap V$ retracts by deformation on two circles, one per piece, while $U,V$ both retract to a single one. Therefore, $U\cap V$ should have cohomology isomorphic to $\mathbb{R}^2$, and $U,V$ should have cohomology isomorphic to $\mathbb{R}$. For the Künneth formula, I can see $U$ and $V$ as homeomorphic to $(-1,1)\times S^1$, so their cohomologies should both be $H^0((-1,1))\otimes H^1(S^1)\oplus H^0(S^1)\otimes H^1((-1,1))$, and since the sets are both path connected the $H^0$ are trivial, $H^1$ of the interval is also trivial since the interval is homeomorphic to $\mathbb{R}$ which should have no cohomology as other euclidean spaces, and $H^1(S^1)$ is $\mathbb{R}$, so the result is $1\otimes\mathbb{R}\oplus1\otimes1\cong\mathbb{R}$. The intersection, however, is $((-1,0)\cup(0,1))\times S^1$, so by Künneth I have the abouve with the interval changed to $(-1,0)\cup(0,1)$, whose cohomology is trivial again, but whose $H^0$ is $\mathbb{R}^2$, since it has two (path) connected components. So I would have cohomology $\mathbb{R}^2$. I must be getting something wrong in this formula, as $U,V$ would end up having the same cohonology as the intersection thinking on it again, since $H^0$ of an interval is not trivial but $\mathbb{R}$, actually I would get $\mathbb{R}^2\oplus\mathbb{R}$, OK let's just ignore Künneth. Intuition anyway says $U,V$ have cohomology $\mathbb{R}$ and their intersection has cohomology $\mathbb{R}^2$, but when I look at possible generators, I find they are cohomologous! Also, if $i_U,i_V$ are the inclusions of the intersection into $U,V$ respectively, what maps do they induce? I mean, if $\alpha,\beta$ are cohomologous, it means $\alpha-\beta=\mathrm{d}f$ for some function $f$, so if they are on $U$, it means $f$ is defined at least on $U$, and so on any subset of $U$, which should mean $\alpha,\beta$ are cohomologous in the intersection $U\cap v$ too. So to get extra cohomology from the intersection I should find a closed, not exact form on the intersection that is not defined on the whole $U$, which implies it must be undefined for some point with $x=0$. But how do I? And are $\omega_1,\omega_2$ actually cohomologous? If they are, how come $\mathbb{R}^2$ minus two points has the two angular forms as distinct generators of its cohomology?
PS By analogous arguments I should find the torus $M$ has cohomology $\mathbb{R}^2$, with generators an angular form such as $\omega_1$ and the angular form around the origin in the $(x,y)$ plane, extended by translational invariance along the $z$ axis.