De Rham Cohomology: tridimensional space $\mathbb{R}^{3}$ without a circle

Question

I'm a new users. I'd like to calculate the De Rham cohomology of euclidean space $\mathbb{R}^{3}$ without a circle $\mathbb{S}^{1}$. I don't have idea how to proceed! I saw the answer given to this question here but I don't understand nothing. The only tools that I know are the Mayer-Vietoris sequence and the equivalence of cohomology of homotopic manifolds. I also know the De Rham cohomology of n-dimensional sphere $\mathbb{S}^{n}$ and the De Rham cohomology of projective spaces. How can I proceed? Should I use Mayer-Vietoris? Or should I find an homotopy between $M=\mathbb{R}^{3}$ \ $\mathbb{S}^{1}$ and another manifold simplier to calculate?

Answer 1

Hint: $\mathbb{R^3}$\ $\mathbb{S^1}$ is deformation retracts to $ \mathbb{S^2} \vee\mathbb{S^1}$ (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: $U \cap V$ 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

Look at the answer by Pedro Tamaroff in the linked question. He does this for homology but it is basically the same.

Say the circle is $C = \{(x,y,z) : x^2 + y^2 = 1, z = 0\}$.

Take $\alpha \in (0,1)$ and define the open cubes $\tilde{U} = \{(x,y,z): x < \alpha\}$ and $\tilde{V} = \{(x,y,z): x > -\alpha\}$. Of course $\tilde{U},\tilde{V}$ are diffeomorphic to $\mathbb{R}^3$.

Define $U := \tilde{U} \cap (\mathbb{R}^3 \setminus C) = \tilde{U} \setminus C$ and $V := \tilde{V} \cap (\mathbb{R}^3 \setminus C) = \tilde{V} \setminus C$

Now observe that both $U$ and $V$ are open and diffeomorphic to $\mathbb{R}^3$ minus a line, and $U \cap V$ is diffeomorphic to $\mathbb{R}^3$ minus 2 distinct lines.

Moreover observe that $\mathbb{R}^3$ minus $n$ distinct lines smoothly retracts on $\mathbb{R}^2$ minus $n$ distinct points. Thus $H^2$ of $U$,$V$ and $U \cap V$ is zero and $H^1(U) \cong H^1(V) \cong \mathbb{R}$, $H^1(U \cap V) = \mathbb{R}^2$.

Now you can apply Mayer-Vietoris:

$$ 0 \to \mathbb{R} \to \mathbb{R} \oplus \mathbb{R} \to \mathbb{R} \to H^1 \to \mathbb{R}\oplus \mathbb{R} \to \mathbb{R}^2 \to H^2 \to 0$$

By analyzing the map $\mathbb{R}^2 \to \mathbb{R}^2$ you get that its image is one dimensional and hence $H^2 \cong \mathbb{R}$.

By dimensionality (the fact that alternating sum of dimensions is zero in an exact sequence) $H^1$ should be also 1-dimensional ($1-2+1+2-2+1 = 1$).