if f and g are two smooth functions from a smooth manifold M^n->S^p,
and we have the condition
that |f-g|<2 then f and g are smoothly homotopic. I think I
understand
that in this case f and g
are both members of a cohomotopy group, but the significance of <2
evades me. Clearly, if
|f-g|>2, then either one of both of the maps don't map onto the n-
sphere, if |f-g|=2, then f,g
map to antipodal points, but don't see right away why that scotches
their being homotopic.
Thanks in advance for any suggestions on the significance of the
inequality condition.
Re: Prove that Right Trivialization is smooth? ... Is there an elegant way to prove that Right Trivialization is smooth?... the Tangent map at point p is denoted by T_g. ... This is a vector bundle isomorphism. ... (sci.math)
Re: Prove that Right Trivialization is smooth? ... Is there an elegant way to prove that Right Trivialization is smooth?... the Tangent map at point p is denoted by T_g. ... I think you mean that a vector bundle is trivial if there are ... (sci.math)
Re: Prove that Right Trivialization is smooth? ... but it needs some basic knowledge in how objects in linear algebra paste together to smooth maps.... the Tangent map at point p is denoted by T_g. ... T_edefines a vector field for each m. ... Next guess the inverse map.... (sci.math)
Re: Prove that Right Trivialization is smooth? ... Is there an elegant way to prove that Right Trivialization is smooth?... the Tangent map at point p is denoted by T_g. ... result like the following product rule on manifolds invoking Phi from ... (sci.math)
cohomotopy result ... if f and g are two smooth functions from a smooth manifold M^n->S^p, ...map to antipodal points, but don't see right away why that scotches ... (sci.math.research)
