Extending smooth maps from $\mathbb{R}^3$ to $S^3$

Question

Suppose I have a smooth map $f\colon \mathbb{R}^3 \longrightarrow S^2$. If I identify $\mathbb{R}^3$ with $U_S = S^3 - \{(0,0,1)\}$ via stereographic projection, $\varphi_S\colon U_S \longrightarrow \mathbb{R}^3$, $\varphi_S(y^1, y^2, y^3, y^4) = \left(\frac{y^1}{1-y^4}, \frac{y^2}{1-y^4}, \frac{y^3}{1-y^4} \right)$, and suppose $f(\vec{y})$ approaches a constant as $\lVert \vec{y}\rVert \rightarrow \infty$, then $f \circ \varphi_S$ defines a map from $U_S$ to $S^2$ that extends via continuity to a map from $S^3$ to $S^2$. My question now is, in what circumstance can we say that this extended map is also smooth?

Answer 1

Let $\imath$ denote inversion, $$ \imath(x, y, z) = \frac{(x, y, z)}{x^{2} + y^{2} + z^{2}},\quad (x, y, z) \neq (0, 0, 0). $$ The extension of $f$ to the three-sphere is smooth if and only if $f\circ\imath$ extends smoothly to the origin.