Skip to main content
added 1 character in body
Source Link
Mehdi Yazdi
  • 858
  • 4
  • 12

You can think about attaching a solid torus as a two-step process:

First, attach $D^2 \times I$ where $D^2$ is the meridional disc of the attaching torus.

Secondly, attach the remaining of the solid torus, which is homeomorphic to a ball $B^3$.

Now if we know where the meridian of the solid torus goes, then we know the result of the first step, $M_1$, up to isotopy. On the other hand, the result of the second step is determined by the map $\phi \colon \partial B^3 \rightarrow \partial M_1$, up to isotopy. By Smale's theorem, any orientation-preserving homeomorphismdiffeomorphism of the 2-sphere is smoothly isotopic to the identity map. Hence the second step is essentially done uniquely.

You can think about attaching a solid torus as a two-step process:

First, attach $D^2 \times I$ where $D^2$ is the meridional disc of the attaching torus.

Secondly, attach the remaining of the solid torus, which is homeomorphic to a ball $B^3$.

Now if we know where the meridian of the solid torus goes, then we know the result of the first step, $M_1$, up to isotopy. On the other hand, the result of the second step is determined by the map $\phi \colon \partial B^3 \rightarrow \partial M_1$, up to isotopy. By Smale's theorem, any orientation-preserving homeomorphism of the 2-sphere is smoothly isotopic to the identity map. Hence the second step is essentially done uniquely.

You can think about attaching a solid torus as a two-step process:

First, attach $D^2 \times I$ where $D^2$ is the meridional disc of the attaching torus.

Secondly, attach the remaining of the solid torus, which is homeomorphic to a ball $B^3$.

Now if we know where the meridian of the solid torus goes, then we know the result of the first step, $M_1$, up to isotopy. On the other hand, the result of the second step is determined by the map $\phi \colon \partial B^3 \rightarrow \partial M_1$, up to isotopy. By Smale's theorem, any orientation-preserving diffeomorphism of the 2-sphere is smoothly isotopic to the identity map. Hence the second step is essentially done uniquely.

Source Link
Mehdi Yazdi
  • 858
  • 4
  • 12

You can think about attaching a solid torus as a two-step process:

First, attach $D^2 \times I$ where $D^2$ is the meridional disc of the attaching torus.

Secondly, attach the remaining of the solid torus, which is homeomorphic to a ball $B^3$.

Now if we know where the meridian of the solid torus goes, then we know the result of the first step, $M_1$, up to isotopy. On the other hand, the result of the second step is determined by the map $\phi \colon \partial B^3 \rightarrow \partial M_1$, up to isotopy. By Smale's theorem, any orientation-preserving homeomorphism of the 2-sphere is smoothly isotopic to the identity map. Hence the second step is essentially done uniquely.