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$\pi:R^3 \to R^2$ given by $ \pi(x,y,z)=(x,y)$

In the textbook, do Carmo-Differential geometry of curves & surfaces, page 66, the proof of proposition 4, it says that the projection $\pi$ restricted on a neighborhood is continuous.

I think that the projection defined by $\pi$ is continuous on $R^3$, and I wonder if condition in the proposition 4: $S$ is a regular surface is necessary

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1 Answer 1

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The restriction of a continuous map to any topological subspace whatsoever is always continuous. The hypotheses are used for the rest of the argument — in particular, to apply the inverse function theorem.

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  • $\begingroup$ Thanks for answering, so to apply the inverse function theorem the $S$ must be regular surface? $\endgroup$
    – Gang men
    Commented Mar 8, 2023 at 5:46
  • $\begingroup$ Absolutely, yes. $\endgroup$
    – Ted Shifrin
    Commented Mar 8, 2023 at 6:13
  • $\begingroup$ Thank you so much and I get it now! $\endgroup$
    – Gang men
    Commented Mar 8, 2023 at 9:34
  • $\begingroup$ By the way, I wonder if the definition of continuous map refers to continuous in $R^3$? Because I consulted my teacher and he told me that the domain of $\pi$ must inherit the toplogy of $R^3$. And it means that the surface $S$ must be regular surface. So I am a a little bit confused that which explanation is true. $\endgroup$
    – Gang men
    Commented Mar 10, 2023 at 9:21

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