# Thom space of unit circle

Say we embed $S^1$ into $\mathbb{R}^2$ as the unit circle. What is the Thom space $Th(i)$ associated to this embedding $i:S^1 \to \mathbb{R}^2$? By definition, the Thom space is the one point compactification of of the normal bundle. I guess I can visualize the normal bundle as homeomorphic to a tubular neighborhood of $S^1$. My first question is basically how can I visualize the one point compactification of this tubular neighborhood, i.e. is there a more familiar topological space that is homeomorphic to $Th(i)$?

In notes I'm reading, the one-point compactification construction is (somewhat surprisingly) contravariant for inclusions such as $i$, i.e we have a map $i^+:\mathbb{R}^2 \coprod * \to Th(i)$, i.e $i^+:S^2 \to Th(i)$, i.e an element of $\pi_2(Th(i))$. In the case at hand, is this element $0$?

• The unit disc bundle of the normal bundle is $S^1 \times [0,1]$. An equivalent definition of the Thom space of a bundle $\nu$ (over a paracompact space) is $D(\nu)/S(\nu)$, where $S$ is the unit sphere bundle. So you obtain $S^1 \times [0,1] / (S^1 \times \{0,1\})$. This, in turn, is homemorphic to a copy of the 2-torus with its inner circle collapsed to a point. (It looks like a sort of singular torus.) – user98602 Jul 5 '15 at 5:48

A tubular neighborhood of $S^1$ in $\mathbb{R}^2$ is an annulus (which is topologically just a cylinder). Making the one-point compactification is just collapsing the top and bottom borders of the cylinder together into a point, which is equivalently a sphere that's had two points collapsed: