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Given an invertible function $f: \mathbb R ^n \rightarrow \mathbb R ^m$, how can we show that it is one to one and invertible?

Given an example such as $f(x,y) = (y + x,\sin y ,yx)$, what can we conclude? If this was a function that took $(x,y) \rightarrow \mathbb R$, it would be enough that the derivative be no zero on its domain, but what is the equivalent statement here? There's also the problem of invertiblity. Since this particular function is not class $C^{\infty}$, we can't call on the inverse function theorem.

Astrum
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  • In this case, injectivity is easy to verify. If $f(x,y)=f(x',y')$, then the third component implies $y=y'$, and then the first component implies $x=x'$. But I suppose you are asking for something more general... – angryavian Sep 28 '16 at 06:27
  • Yeah, it was a bad example, I'll edit it. – Astrum Sep 28 '16 at 06:49

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