Is there a manifold $M\subset\mathbb{R}^3$ s.t. $H^k(M)=\mathbb{R}$ for $k=1,2,3$? (deRham cohomology groups)

Question

I was thinking the following: First, note that $\mathbb{S}^1\subset\mathbb{R}^2$ and we have the deRham cohomology groups $H^{k}(\mathbb{S}^1)=\mathbb{R}$ for $0\leq k<2$ and $0$ otherwise, but $\mathbb{S^1}$ is not embedded on $\mathbb{R}$ ($\mathbb{S}^1$ is of dimension one). This motivates a curiosity: Is there a manifold $M\subset\mathbb{R}^3$ such that $H^{k}(M)=\mathbb{R}$ for $k=1,2,3$? (In analogy to the case of $\mathbb{S}^1$ but now embedded in a space of the same dimension of its dimension) Of course such manifold must be of dimension 3, but I can't imagine the shape and topology it must be. Anyone know such example of manifold?

Answer 1

Let $X = S^1 ⨆S^2 ⨆ S^3 \subset \mathbb R^4$. Then, clearly $H^k(X)=\mathbb R$ for $k=1,2,3$.

It is not possible to achieve this as a subspace topology of $\mathbb R^3$. That is to say that a three dimensional object cannot have a four dimensional hole. Existence of such an object would give rise to an injection/embedding $S^3 \to \mathbb R^3$, which is impossible.

Note: ⨆ here is the disjoint union of topological spaces.