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?
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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. |
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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. |
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