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Pictures below are from the Do Carmo's Riemannian Geometry. He want to show $i$ is a embedding. But only at $U$, $i$ is homeomorphism since $j^{-1}\circ i \circ x=x$. This do not mean $i$ is homeomorphism on $S$. Since in my view, the embedding means that $i:S\rightarrow i(S)$ is homeomorphism. However, only local homeomorphism is proved in picture below. So, this proof don't prove the $i$ is embedding. How to deal it ?

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Enhao Lan
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  • The inclusion map is very clearly a bijection onto its image. But a continuous bijection which is a local homeomorphism is actually a global homeomorphism (it is not hard to check that $i^{-1}$ is continuous using the local homeomorphism property). – Glare Dec 27 '20 at 12:57

1 Answers1

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Stating that the inclusion $i \colon S \longrightarrow \mathbb{R}^3$ is a homeomorphism onto its image is the same as saying that, considering the inclusion now as being the map $i \colon S \longrightarrow i(S) = S \subset \mathbb{R}^3$, we get a homeomorphism. But that is obvious, since now it becomes the identity on $S$ and $S$ is equipped with the same topology (the subspace topology) as both the domain and the image of $i$ .

So, actually, the fact that $i$ is a homeomorphism onto its image does not "follow from a)", but it is trivial.

Hope this helps :).

Duarte Costa
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