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I'm trying to work through the following problem.

Let $M$ be an $m$-dimensional manifold. Prove that the graph of any smooth map $f:M\to\mathbb{R}$ is a closed, embedded submanifold of $M\times\mathbb{R}$.

My Attempt (so far): We recall the Regular Value Theorem.

If $F:M^{m}\to N^{n}$ is a smooth map and $q\in N$ is a regular value of $F$ (i.e. the differential $DF$ is onto for each $p\in f^{-1}(q)$), then $F^{-1}(q)$ is a closed, embedded submanifold of $M$.

Let $\Gamma_{f}=\{(p,q)\in M\times\mathbb{R}~:~q=f(p)\}$ be the graph of the smooth map $f:M\to\mathbb{R}$. Define $\phi:M\times\mathbb{R}\to\mathbb{R}$ by $\phi(p,q)=q$. Then $\phi$ is smooth and $$\phi^{-1}(f(p))=\{(p,q)\in M\times\mathbb{R}~:~\phi(p,q)=q=f(p)\}=\Gamma_{f}.$$ To show that $\Gamma_{f}$ is a closed, embedded submanifold of $M\times\mathbb{R}$, it will suffice (by the Regular Value Theorem) to show that $f(p)$ is a regular value of $\phi$. Hence, we must show that the differential $D\phi$ is onto for each point in $\phi^{-1}(f(p))$.

My Questions/Concerns: Does this argument look okay so far? If it does, my only concern is I'm not sure how to show that the differential is surjective.

Thanks in advance for any suggestions!

Sir_Math_Cat
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  • Is $M$ connected? – Baol Sep 30 '18 at 22:00
  • @Baol It wasn't given as a requirement. – Sir_Math_Cat Sep 30 '18 at 22:03
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    Consider $\pi:M \times \mathbb{R} \rightarrow M$, $\phi:M \rightarrow M \times \mathbb{R} :p \mapsto (p,f(p))$ then $\pi \circ \phi=id_M$ so its tangent mapping is the identity, and thus $T_p\phi$ is always injective (and smooth) so it is a smooth immersion. Then finally you provide a local inverse with the projection $\pi$. So $\phi$ becomes an embedding. – Baol Sep 30 '18 at 22:24
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    This is related, I hope it is useful to you. (I am also a coffee addict, by the way). – Giuseppe Negro Oct 01 '18 at 09:23

1 Answers1

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Consider the maps $\pi:M\times \mathbb{R} \rightarrow M: (p,t) \mapsto p$, and $\phi:M \rightarrow M\times \mathbb{R}: p \mapsto (p,\phi(p))$. Clearly $\pi \circ \phi = id_M$, so if we take the tangent map (alternative name for the differential) we have $T_{(p,\phi(p))}\pi \circ T_p\phi=T_p id_M=id_{T_p M}$, form this one sees that $T_p\phi$ must be injective for every $p$, hence $\phi$ is an immersion. To proof that it is an embedding we have to provide a local inverse which is continous, then $\phi$ will become a local homeomorphism onto its image, to see this, on $\Gamma_f$ one has that $\phi \circ \pi\mid_{\Gamma_f} = id_{\Gamma_f}$ moreover $\pi$ is continous almost by definition of product topology, so a conclusion is reached.

Baol
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  • Baol, what is a local diffeomorphism onto image exactly please? I know what local diffeomorphisms, local homeomorphisms onto image and local homeomorphisms are. The issue with copying the definition of local homeomorphism onto image to local diffeomorphism onto image has the problem of the image possibly not being a submanifold or manifold while there is no such issue for local homeomorphism onto image since image can always be made into a subspace. –  Jul 22 '19 at 05:55