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Is the smoothness required in the following question ? I think continuity is sufficient to prove the function being continuous.

enter image description here

My attempt:

As $g$ is a smooth function , $g$ will also be a continuous function.

Now if $g$ was not a constant function , it would have taken two values $1 , -1$. So $g^{-1}(-1 )$ and $g^{-1}(1 )$ are two disjoint open sets of $S$. Besides , $g^{-1}(1) \cup g^{-1}(-1) = S$ as the co-domain set of $g$ contains only $1 , -1$.

So $S$ is not connected.

But it is given that $S$ is a connected surface of $\mathbb R^n$.

Therefore , $g$ is a constant function.

cmi
  • 3,371
  • I think I agree. (Also weird to phrase the question as "tak[ing] on only" the values $1$ and $-1$, when they mean has range contained in ${-1, 1}$. I mean, it doesn't take both values on, does it?) See e.g. https://math.stackexchange.com/questions/1573795/proof-of-the-continuous-image-of-a-connected-set-is-connected – leslie townes Feb 27 '21 at 06:38
  • Yes I know. That is the point. As $S$ is a connected surface , it will be a constant function.@leslietownes – cmi Feb 27 '21 at 07:44
  • please check my attempt@leslietownes – cmi Feb 27 '21 at 07:53
  • looks good to me. – leslie townes Feb 27 '21 at 07:56

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