# Smooth structure of embedded submanifolds

Here is a rather long quotation from "Introduction to smooth manifolds" by J. Lee.

Proposition 5.2 (Images of Embeddings as Submanifolds). Suppose $M$ is a smooth manifold with or without boundary, $N$ is a smooth manifold, and $F: N \to M$ is a smooth embedding. Let $S = F(N)$. With the subspace topology, $S$ is a topological manifold, and it has a unique smooth structure making it into an embedded submanifold of $M$ with the property that $F$ is a diffeomorphism onto its image.

Proof. If we give $S$ the subspace topology that it inherits from $M$, then the assumption that $F$ is an embedding means that $F$ can be considered as a homeomorphism from $N$ onto $S$, and thus $S$ is a topological manifold. We give S a smooth structure by taking the smooth charts to be those of the form $(F(U), \phi \circ F^{-1})$, where $(U, \phi)$ is any smooth chart for $N$; smooth compatibility of these charts follows immediately from the smooth compatibility of the corresponding charts for $N$. With this smooth structure on $S$, the map $F$ is a diffeomorphism onto its image (essentially by definition), and this is obviously the only smooth structure with this property. ...

My question is about the bold sentence. He says that it is obvious, but it is not obvious to me. Why is the constructed smooth structure the only smooth structure that makes $F$ a diffeomorphism between $N$ and $S$?

• If $\widetilde S$ represents the topological manifold $S$ endowed with some smooth structure such that $F$ is a diffeomorphism onto its image, then for any smooth chart $(U,\phi)$ for $N$, it follows that $(F(U),\phi\circ F^{-1})$ will be a smooth chart for $\widetilde S$ because $F(U)$ is open in $S$ and $\phi\circ F^{-1}$ is a composition of diffeomorphisms. So the unknown smooth structure on $S$ must be the one determined by these charts. – Jack Lee Feb 14 '17 at 1:16
• @JackLee I didn't expect a response from the author! Thank you very much. Now I understand it. – leeto Feb 14 '17 at 14:13
• You're welcome! – Jack Lee Feb 14 '17 at 15:30

## 1 Answer

If $\widetilde S$ represents the topological manifold $S$ endowed with some smooth structure such that $F$ is a diffeomorphism onto its image, then for any smooth chart $(U,\phi)$ for $N$, it follows that $(F(U),\phi\circ F^{−1})$ will be a smooth chart for $\widetilde S$ because $F(U)$ is open in $S$ and $\phi\circ F^{-1}$ is a composition of diffeomorphisms. So the unknown smooth structure on $S$ must be the one determined by these charts.