differential geometry - Why is $rm{Conf}(mathbb{R}^{1,1}) = rm{Diff}(S^1) times rm{Diff}(S^1)$ and not $ rm{Diff}(mathbb{R}) times rm{Diff}(mathbb{R})$?

The Minkowski metric for $\mathbb{R}^{1,1}$ is $$ ds^2 = dt^2 - dx^2 = du dv $$ for coordinates $$ u = t + x \hspace{1cm} v = t - x $$

If you do any coordinate transformation that acts independently one $u$ and $v$,

$$ u = f(u') \hspace{1 cm} v = g(v') $$ then the metric transforms as $$ ds^2 = du dv = d(f(u')) d(f(v')) = f'(u') g'(v') du' dv'. $$

Notice that the metric in $(u', v')$ coordinates is just a local rescaling of the original metric $dudv$. Therefore this is a conformal transformation. In fact, all conformal transformations of $\mathbb{R}^{1,1}$ are of this form.

Why then do I often see written $\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(S^1) \times \rm{Diff}(S^1)$? Shouldn't it just be $\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(\mathbb{R}) \times \rm{Diff}(\mathbb{R})$? Why is $\mathbb{R}$ "compactified" into $S^1$?