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!
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!
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.
Again, this is just an elaboration of Mariano's answer.
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.