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Hi:

This is , I guess a technical question:

Given f:R^m --->R^m ; f=(f_1(x_1,..,x_m),....,f_m(x_1,...,x_m))

Then I guess f^-1 (of course, assume f is 1-1.). Is given by a "pointwise" inverse ,

(right?) i.e.,

f^-1 =(f_1^-1 (x_1,..,x_m) ,....,f_m^-1(x_1,..,x_m)) ?.

Is there some theorem on existence of inverses if we only know f to be

continuous ( I think there is no known test for whether a function into R^m

is onto )?

Thanks.

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# Inverse functions for f:R^m->R^m , or f:X^m->Y^m

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