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As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a two-orbifold (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation).

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. The only such orbifold is $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$. But the fundamental group of any such bundle is $\mathbb{Z}$, contradicting the fact that $\mathbb{R}^3$ is simply connected.

Smoothness is not really necessary. Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibering of a solid torus". If this holds, then the partition gives $\mathbb{R}^3$ the structure of a Seifert fibered space. The non-compact case is more difficult than the compact case, but there is still a rich theory.

The extra structure (of namely having good neighbourhoods) holds when the parts of the partition are (continuously varying) round circles. So the answer to Question 1 is no. Question 2 feels more subtle; after all, continuous loops in space can be pretty badly behaved! So obtaining the bundle structure seems more difficult.

As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a two-orbifold (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation).

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. The only such orbifold is $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$.

Now, $\mathbb{R}^2$ is contractible. So any $S^1$-bundle over it is isomorphic to $S^1 \times \mathbb{R}^2$. This has fundamental group $\mathbb{Z}$ and we are done.

Smoothness is not really necessary. Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibering of a solid torus". If this holds, then the partition gives $\mathbb{R}^3$ the structure of a Seifert fibered space. The non-compact case is more difficult than the compact case, but there is still a rich theory.

The extra structure (of namely having good neighbourhoods) holds when the parts of the partition are (continuously varying) round circles. So the answer to Question 1 is no. Question 2 feels more subtle; after all, continuous loops in space can be pretty badly behaved! So obtaining the bundle structure seems more difficult.

As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a two-orbifold (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation).

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. The only such orbifold is $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$. But the fundamental group of any such bundle is $\mathbb{Z}$, contradicting the fact that $\mathbb{R}^3$ is simply connected.

Smoothness is not strictly necessary. Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibring of a solid torus". This obtained, the partition gives $\mathbb{R}^3$ the structure of a Seifert fibered space. The non-compact case is more difficult than the compact case, but there is still a rich theory.

As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a two-orbifold (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation).

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. The only such orbifold is $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$. But the fundamental group of any such bundle is $\mathbb{Z}$, contradicting the fact that $\mathbb{R}^3$ is simply connected.

Smoothness is not really necessary. Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibering of a solid torus". If this holds, then the partition gives $\mathbb{R}^3$ the structure of a Seifert fibered space. The non-compact case is more difficult than the compact case, but there is still a rich theory.

The extra structure (of namely having good neighbourhoods) holds when the parts of the partition are (continuously varying) round circles. So the answer to Question 1 is no. Question 2 feels more subtle; after all, continuous loops in space can be pretty badly behaved! So obtaining the bundle structure seems more difficult.

As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a surface. Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. So the quotient space is homeomorphic to $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$. But the fundamental group of any such bundle is $\mathbb{Z}$, contradicting the fact that $\mathbb{R}^3$ is simply connected.

As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a two-orbifold (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation).

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. The only such orbifold is $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$. But the fundamental group of any such bundle is $\mathbb{Z}$, contradicting the fact that $\mathbb{R}^3$ is simply connected.

Smoothness is not strictly necessary. Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibring of a solid torus". This obtained, the partition gives $\mathbb{R}^3$ the structure of a Seifert fibered space. The non-compact case is more difficult than the compact case, but there is still a rich theory.