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I need to show that every smooth vector field along an immersion has local smooth extensions.

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 Your question is too vague and ill-formed. I suggest reading MO for a while before making your first post, to get a sense for what is expected from question-askers. Also, reading the FAQ helps if you have not yet. – Ryan Budney May 9 2011 at 0:25

locked by S. Carnahan♦ May 25 2011 at 3:36

closed as too localized by Will Jagy, Ryan Budney, Deane Yang, Charles Siegel, Willie Wong May 9 2011 at 3:13

2 Answers

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If you do not demand non-vanishing, there is never a problem to extend a local section of a vector bundle to a global section. Use local trivializations to extend them locally and glue these together with a partition of unity.

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Consider such an immersed curve curve (without endpoints!) and its tangent vector field. This field cannot have smooth extension around the points where the curve is almost touching itself.

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 @r0b0t Isn't the tangent vector field a function from the image set into $R^2$? If so, then it is not even continuous at the two interior points where the "ends" approach the image, since the tangent vectors at those points are not the limits of the tangent vectors along the approaching ends. – Dick Palais May 9 2011 at 0:29
The only definition of "smooth along an immersion" I see is that the pullback is smooth. In the continuous category I can imagine also another definition as "being continuous on the image". For immersed manifolds which are not embedding these notions clearly differ. How would you define a smooth function, when the subset is not a submanifold? – robot May 9 2011 at 10:07
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 You could consider the ring $C^\infty(\mathbb R^2)$ modulo the ideal of smooth functions that vanish on the image of your set. You could then define a "vector field" to be a derivation of this algebra. In the case of embedded submanifolds, this definition also agrees with the various others. – Theo Johnson-Freyd May 25 2011 at 2:04

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