Differential structure on regular submanifold vs. embedded submanifold

Question

Let $M$ and $N$ be smooth manifolds and let $F:M\longrightarrow N$ be a smooth embedding. Here the definition of smooth embedding is as follows: $$F:M\rightarrow N \;\text{is a one-to-one immersion, and }F:M\rightarrow F(M)\;\text{is a homeomorphism with subspace topology on }F(M)\subset N.$$

1. Then we know that $F(M)$ is an embedded submanifold of $N$ with subspace topology. If $\Sigma=\{(U_{\alpha},\varphi_{\alpha}) \}$ is an atlas on $M$, then we know $\{(F(U_{\alpha}),\varphi_{\alpha}\circ F^{-1}\vert_{F(U_{\alpha})} )\}$ is an atlas on $F(M)$, which determines a differential structure on $F(M)$. We denote this differential structure by $DS_{1}$.
2. Also, we know that $F(M)$ can be viewed as a regular submanifold of $N$ with subspace topology. And for each point $p\in F(M)$, there exists a chart $(U,\varphi)$ on $N$, such that $\varphi(U\cap F(M))=\varphi(U)\cap \left(\mathbb{R}^{k} \times \{0\} \right)$. So we know $(U\cap F(M),\varphi |_{U\cap F(M)})$ is a chart around $p$ for this regular submanifold $F(M)$. And this also provides a differential structure on $F(M)$, denoting by $DS_{2}$.

My confusion is: Must we have $DS_{1}=DS_{2}$? Can anyone help me?

================================================================================================= Update ================================ My professor told me a proposition that may work for this problem.

Proposition: Let $(M,\tau)$ be a smooth manifold. $\Sigma_{1}$ and $\Sigma_{2}$ are two smooth atlases on $M$. The differential structures they determines denote by $DS_{1}$ and $DS_{2}$ respectively. Then we have $DS_{1}=DS_{2}\Leftrightarrow \Sigma_{1}\cup \Sigma_{2}$ is still a smooth atlas on $M$.

I have tried to use this proposition to solve my problem, as shown in figure.

The only thing I need to check is the smoothness of the transition maps $\phi |_{V\cap F(M)}\circ (\varphi\circ F^{-1})^{-1}$ and $(\varphi\circ F^{-1})\circ\phi^{-1}$. How can I know these two maps are smooth?

Answer 1

Yes, they are the same. Let $X$ denote $F(M)$ with the smooth structure given by $DS_1$ and let $Y$ denote $F(M)$ with the smooth structure given by $DS_2$. Note first that if $f:P\to N$ is any smooth map from a manifold $P$ whose image is contained in $F(M)$, then $f$ is smoooth as a map to $P\to Y$. Indeed, just look in a local chart on $N$ as in the definition of $DS_2$: locally, $f$ looks like a smooth map $\mathbb{R}^p\to\mathbb{R}^n$ whose image happens to be contained in $\mathbb{R}^k\times\{0\}$, and such a map is obviously also smooth when considered as a map to $\mathbb{R}^k$ (since you are just taking the first $k$ coordinates of the map).

Applying this to the inclusion map $X\to M$, we see that the identity map $i:X\to Y$ is smooth. But also, since $F$ was an immersion, $di$ is injective everywhere (restricting the codomain to $Y$ does not change that, since looking locally and identifying $N=\mathbb{R}^n$, $Y=\mathbb{R}^k\times\{0\}$, the derivatives of all coordinates of $F$ after the $k$th are identically $0$, so if $dF$ is injective it must still be injective looking at just the first $k$ coordinates). Note moreover that $X$ and $Y$ have the same dimension: by the constant rank theorem, $i:X\to Y$ looks locally like the inclusion $\mathbb{R}^{\dim X}\times\{0\}\to\mathbb{R}^{\dim Y}$. But since $F$ was an embedding, $i$ is a homeomorphism and in particular it is locally an open map, and the inclusion $\mathbb{R}^{\dim X}\times\{0\}\to\mathbb{R}^{\dim Y}$ is only an open map if $\dim X=\dim Y$.

Finally, the inverse function theorem now says that $i^{-1}$ is smooth, since $di$ is an isomorphism everywhere. Thus $i:X\to Y$ is a diffeomorphism, which means the two smooth structures $DS_1$ and $DS_2$ are the same.