Suppose $v\colon \mathbb{R}^3\setminus\{x_1,\dots,x_n\}\to S^2$ is a smooth map, where all the points $x_i$ sit inside the unit ball $B^3$, and not on the boundary $S^2$. I suppose this map can be thought of as a unit vector field.

If we restrict $v|_{S^2}:S^2\to S^2$, we have a smooth map on continuous, compact, oriented $2$-manifolds, so $\deg(v|_{S^2})=\sum_{x\in v|_{S^2}^{-1}(q)}\operatorname{sgn}(x)$ where $q$ is a regular value, and $\operatorname{sgn}(x)=\pm1$ depending on whether $dF_x$ is orientation preserving/reversing.

What is the reason that this degree actually coincides with the sum of indices of $v$ at the deleted points $x_i$?


Because each index is itself the degree on a tiny sphere centered at the point. Moreover, if $f\colon W\to Y$ is a smooth map, $W$ a compact $(n+1)$-manifold with boundary, $Y$ a compact $n$-manifold, both oriented, then $\text{deg}(f|_{\partial W}\colon\partial W\to Y)=0$.

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  • $\begingroup$ Thanks, so are you viewing $v:B^3\to S^2$ as a smooth map, so that $v|_{S^2}$ has degree $0$? I found a proof of this statement as Theorem 17.38 in the book Intro to Smooth Manifolds by Lee, but how do you know the sum of the indices on the small spheres around the $x_i$ add up to $0$ as well? $\endgroup$ – Morneau Sep 20 '14 at 6:28
  • $\begingroup$ You apply my statement to $W=B^3-\cup B(x_i,\epsilon)$. Note that there's an orientation reversal on $\partial B(x_i,\epsilon)$. $\endgroup$ – Ted Shifrin Sep 20 '14 at 12:48
  • $\begingroup$ Sorry, but if we delete closed balls, isn't $W$ no longer closed, hence not compact? And if we delete open balls, wouldn't that make $\partial W$ disconnected? The Theorem I cited requires $\partial W$ to be connected, but maybe it's not necessary? $\endgroup$ – Morneau Sep 20 '14 at 20:38
  • $\begingroup$ I should have written $\bar B{}^3$, then delete open balls. No, in dimensions $\ge 2$, $W$ will be connected. (This is the standard generalization of the proof of the Cauchy integral formula, residue theorem, etc., in complex analysis.) $\endgroup$ – Ted Shifrin Sep 21 '14 at 1:52
  • $\begingroup$ I see $W$ is conneccted, but wouldn't the boundary $\partial W$ be disconnected since it's a sphere with a bunch of smaller spheres inside it? $\endgroup$ – Morneau Sep 21 '14 at 2:18

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