cohomotopy result
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.
.
Relevant Pages
- 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) |
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