De Rham cohomology of $\mathbb{R}^3$ minus a circumference

Question

Let $M=\mathbb{R}^3-\{\left(x,y,z \right) \ | \ x^2+y^2=1 ,\ \ z=0 \}$. I'd like to calculate its De Rham cohomology.

I've tried to use Mayer Vietoris sequence applied with open sets $U=M-\{x=0 \ \ y=0\}$ and $V$ the open cilynder with axis parallel to the $z$ axis and radius$= \dfrac{1}{2}$, because we know how to calculate the cohomology of $U,V U \cap V$, but this did not really make me end the problem.

Answer 1

Hint: $\mathbb{R^3}$\ $\mathbb{S^1}$ is deformation retracts to $\mathbb{S^1} \vee \mathbb{S^2}$ (A circle wedge sum with a sphere) and now you can use Mayer–Vietoris sequence to decompose the space into two open sets with nonempty intersection as follows: Consider $U=$ The space by removing one point from the circle and $V=$ The space by removing a point from the sphere. Since singleton is closed in a Hausdorff space then the sets $U$ and $V$ are open. Note: $U∩V$ deformation retracts to a point, $U$ is same as the sphere and $V$is same as the circle up to continuous deformation or upto homotopy). I hope this will help. I think the main crux of this problem is to understand the very first statement. For reference please look at Hatcher's Page 46 Example 1.23.

I hope this will help.

Answer 2

To make things a bit easier to see, consider $X = D^3\smallsetminus S^1$, and consider the two open sets $U,V$ obtained by taking two half cubes that intersect slightly in the middle. Then $U$ and $V$ are the same as $D^3$ minus a line, which is $S^1$, and $U\cap V$ is the same as $D^3$ minus two disjoint lines, which is $S^1\vee S^1$.

Since all these spaces are path-connected, you can look at the reduced long exact sequence in homology. Observe that $U,V$ and $U\cap V$ have trivial homology in degrees $>1$, so you get

$$0\to H_2(X)\to H_1(U\cap V)=\mathbb Z^2\to \mathbb Z^2 =H_1(U)\oplus H_1(V) \to H_1(X) \to 0 $$

You can choose generators of $a,b$ of $ H_1(U\cap V)$ that both map to generators of $z$ of $H_1(U)$ and $w$ $H_1(V)$. One, say $a$, will map to $z$ and $w$, and the other will map $-z$ to $-w$. Then the middle map has matrix $$\pmatrix{ 1& 1 \\ -1&-1}$$ The kernel of this matrix is generated by $(1,-1)$ which generates a subgroup isomorphic to $\mathbb Z$, and the image of the map is generated by the same element, so $H_1(X) = H_2(X) =\mathbb Z$. All other (reduced) homology groups vanish.