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

[WWGD]

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

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.

 

[WWGD]

Correction:
Conditions for f should have been (the equiv.) f_1,...,f_n are 1-1.
Thanks.

 

[yyat]

Can you give an example of a continuous function R^m->R (like you f_i) that is 1-to-1? I am not sure such a function exists for m>1.

 

[WWGD]

Yes, it seems difficult, I will think about it. I also realized that the condition that the f_i
are 1-1 is not equivalent to f:R^m-->R^m being 1-1; f :(x,y) = (Pi_1(x,y),Pi_2(x,y) )=
(x,y), with Pi_1, Pi_2 the projections is a counterexample. Life seems to become weird
outside of the safety of functions f:R-->R. Ontoness on each variable does not guarantee
ontoness of f:R^m-->R^m , with the dramatic example of f:R^2-->R^2 :
f_1(x,y) =x =f_2(x,y).

P.S: Sorry for the wordiness of my last post on bdry. of a manifold.

 

[WWGD]

I was wrong again here : If we have f:R^n-->R^n (or f:X^n -->Y^n )

and _just one_ of the f_i's is 1-1 , then f is 1-1 f(x_1,..,x_n)=f(y_1,..,y_n) , (x_1=(x_1i,

etc.)then f_i(x_i)=f_i(y_i) , which cannot happen if f_i is 1-1. Rest is still tricky,

but possible: there are continuous bijections f:X-->Y that are not homeomorphisms

( I need to look it up in my 'Counterexamples in Topology' book).

These product spaces can be confusing (at least to me).

 

[WWGD]

Hi, yyat:

I remembered the answer to your followup on existence of continuous functions
f:R^m-->R :

the answer is no, by k-connectivity: if h:R^m -->R is continuous, then so is

h': R^m-{x} -->R-{h(x)} . But R^m-{x} is connected, and R-{h(x)} is not.

I don't think that there is a nice answer for functions f:R^m-->R^k without

using algebraic topology , i.e., fundamental groups. I will think about this, please

let me know if you come up with some ideas.


Another thought about ontoness of maps f:R^m-->R^m ; maybe not too deep,

but I still think useful: given f=(f_1,..,f_n) , the functions f_i cannot be linearly

dependent, e.g., if f_1(x,y)=x and f_2(x,y) =nx , then f will only hit points in the

(linear) subspace (x,nx), and will miss every point (x+a,nx+b) for (a,b)=/(0,0) ; a,b in R.

This is necessary, together with the ontoness of each f_i ( if , say, f_k misses

y_k , then f will miss RxR..xRx{y_k}xRx..xR ) , but I don't know if these two conditions

together are sufficient to guarantee ontoness.