hence via a standard Stokes’ theorem argument, the “point-splitting regularization integral” used by [@Tze] is the same as the integral along $\mathbb{L}^{\overrightarrow{n}}_{0}(S^{1})$.

### “$C$-swaddling”

Let us eliminate the boundary of our configuration space this time in a different way— rather than pasting two discs onto the $S^{1}\times I$ boundary, let us instead paste another $S^{1}\times I$ to it (which we shall call $C$ for cylinder), completing the cylinder $C_{2}(S^{1})$ to a torus $C_{2}(S^{1})\cup_{\partial}C$ as shown in the illustration below:

$$\xy
0;/r3pc/: (-2,1)="a"; (-1,1)="b" **\crv{(-1.5,1.2)}
**\crv{(-1.5,0.8)}, (-2,0)="c" **@{-}; (-1,0)="d"
**\crv{~*=<3pt>{.}(-1.5,.2)} **\crv{(-1.5,-.2)}; "b" **@{-};
(0,1)="e"; (1,1)="f" **\crv{(.5,1.2)} **\crv{(.5,0.8)}, (0,0)="g"
**@{-}; (1,0)="h" **\crv{~*=<3pt>{.}(.5,.2)} **\crv{(.5,-.2)}; "f"
**@{-}; "a" **\crv{~*=<5pt>{.}(-.5, 2.1)};
 (-1,1)="b"; "e"
**\crv{~*=<5pt>{.}(-.5,1.3)}; (-2,0)="c"; "h"
**\crv{~*=<5pt>{.}(-.5,-1.1)}; (-1,0)="d"; "g"
**\crv{~*=<5pt>{.}(-.5,-.3)} {\ar@{}|{C_{2}(S^{1})}"a";(-1,0)}
{\ar@{}|{C}"e";(1,0)}
\endxy$$  

Then we define a new swaddling map $\phi:C\rightarrow S^{2}$ which maps the boundaries of $C$ to the curves $\pm\dot{\gamma}$ in such a way that $\overline{\Phi_{H}}$ and $\phi$ combine to give us a continuous map $\overline{C_{2}(S^{1})}\cup_{\partial}C\rightarrow
S^{2}$.

As the title to this section suggests, we would like to show that in 3.1.1 and in 3.1.2 we have done one and the same thing (modulo $4\pi$)— that in point of fact, there is no difference between compactifying the space as in section 3.1.1. by 