# Degree of maps on the sphere

Given a map $f:B^n \to S^n$, where $B^n$ is the unit ball and $S^n$ is the unit sphere, is it true that the degree of $f|_{S^n}$ is always 0, where $f_{S^n}$ is the restriction of $f$ to $S^n$? If so, why?

Thanks!

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Do you perhaps want that $f$ is continuous? And also, just by way of clarification: is $S^n$ the boundary of $B^n$ (I personally would write it $S^{n-1}$, but it is just a matter of convention...) –  Willie Wong Jan 3 '11 at 16:59
Yes for both questions, –  user5352 Jan 3 '11 at 17:29
It's not just convention, the superscripts denote the dimension of the manifold! Which makes it really, really confusing to write anything besides $\partial(B^n)=S^{n-1}$... –  Aaron Mazel-Gee Jan 4 '11 at 21:23

This is just a bit more of an explanation of Mariano's answer: The homotopy class of a map between spheres of the same dimension is completely determined by its degree, that is, two maps $f,g: S^n \to S^n$ are homotopic if and only if they have the same degree. Now you can compute, using which ever definition you like, that the degree of the constant map is 0. A map out of a sphere is null homotopic, ie. homotopic to a constant map, if and only if it can be extended to a map of the whole unit ball. Since the map in question, $f|_{S^n}$, can indeed be extended to a map on the whole unit ball, namely $f$, it must be null homotopic. So by the above it must have the same degree as the constant map.
The degree is zero because the restriction of $f$ to the boundary of $B^n$ (which most humans write $S^{n-1}$! :) ) is homotopic to to a constant map.