Proving that the image of an injective, proper immersion is a manifold
I am trying to get through the proof of the statement "if $f: M to N$ is injective, proper and an immersion, then $f:M to f(M)$ is a diffeomorphism onto a submanifold".
The proof I'm reading says that on small sets $U subset M$, $f|_{U}: U to f(U)$ is a diffeomorphism onto its image, using the fact that for immersions $f: M to N$ we can find coordinate charts around $x$ and $f(x)$ such that $f(x_{1},dots,x_{m})=(x_{1},dots,x_{n},0,dots,0)$. However, I'm not sure how this fact about immersions would lead to $f|_{U}$ being a diffeomorphism for small sets U. I think I'm missing something basic but I don't know what it is.
Then, it says that if $f(U)$ is an open set of $f(M)$, then the inverse function is smooth, since it is continuous as $f$ is an open map (I understand this) and then that it is smooth by the inverse function theorem. I don't know how the inverse function theorem would give us this result (it says that for $f: M to N$ smooth then if $Df_{x}$ is an isomorphism of tangent spaces then $f$ is locally a diffeomorphism). Thanks for the help.
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 11 '18 at 15:23
atrenet
61
add a comment |
I am trying to get through the proof of the statement "if $f: M to N$ is injective, proper and an immersion, then $f:M to f(M)$ is a diffeomorphism onto a submanifold".
The proof I'm reading says that on small sets $U subset M$, $f|_{U}: U to f(U)$ is a diffeomorphism onto its image, using the fact that for immersions $f: M to N$ we can find coordinate charts around $x$ and $f(x)$ such that $f(x_{1},dots,x_{m})=(x_{1},dots,x_{n},0,dots,0)$. However, I'm not sure how this fact about immersions would lead to $f|_{U}$ being a diffeomorphism for small sets U. I think I'm missing something basic but I don't know what it is.
Then, it says that if $f(U)$ is an open set of $f(M)$, then the inverse function is smooth, since it is continuous as $f$ is an open map (I understand this) and then that it is smooth by the inverse function theorem. I don't know how the inverse function theorem would give us this result (it says that for $f: M to N$ smooth then if $Df_{x}$ is an isomorphism of tangent spaces then $f$ is locally a diffeomorphism). Thanks for the help.
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 11 '18 at 15:23
atrenet
61
add a comment |
I am trying to get through the proof of the statement "if $f: M to N$ is injective, proper and an immersion, then $f:M to f(M)$ is a diffeomorphism onto a submanifold".
The proof I'm reading says that on small sets $U subset M$, $f|_{U}: U to f(U)$ is a diffeomorphism onto its image, using the fact that for immersions $f: M to N$ we can find coordinate charts around $x$ and $f(x)$ such that $f(x_{1},dots,x_{m})=(x_{1},dots,x_{n},0,dots,0)$. However, I'm not sure how this fact about immersions would lead to $f|_{U}$ being a diffeomorphism for small sets U. I think I'm missing something basic but I don't know what it is.
Then, it says that if $f(U)$ is an open set of $f(M)$, then the inverse function is smooth, since it is continuous as $f$ is an open map (I understand this) and then that it is smooth by the inverse function theorem. I don't know how the inverse function theorem would give us this result (it says that for $f: M to N$ smooth then if $Df_{x}$ is an isomorphism of tangent spaces then $f$ is locally a diffeomorphism). Thanks for the help.
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 11 '18 at 15:23
atrenet
61
I am trying to get through the proof of the statement "if $f: M to N$ is injective, proper and an immersion, then $f:M to f(M)$ is a diffeomorphism onto a submanifold".
The proof I'm reading says that on small sets $U subset M$, $f|_{U}: U to f(U)$ is a diffeomorphism onto its image, using the fact that for immersions $f: M to N$ we can find coordinate charts around $x$ and $f(x)$ such that $f(x_{1},dots,x_{m})=(x_{1},dots,x_{n},0,dots,0)$. However, I'm not sure how this fact about immersions would lead to $f|_{U}$ being a diffeomorphism for small sets U. I think I'm missing something basic but I don't know what it is.
Then, it says that if $f(U)$ is an open set of $f(M)$, then the inverse function is smooth, since it is continuous as $f$ is an open map (I understand this) and then that it is smooth by the inverse function theorem. I don't know how the inverse function theorem would give us this result (it says that for $f: M to N$ smooth then if $Df_{x}$ is an isomorphism of tangent spaces then $f$ is locally a diffeomorphism). Thanks for the help.
differential-geometry differential-topology smooth-manifolds smooth-functions
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 11 '18 at 15:23
atrenet
61
asked Dec 11 '18 at 15:23
atrenet
61
asked Dec 11 '18 at 15:23
atrenet
61
asked Dec 11 '18 at 15:23
atrenet
61
asked Dec 11 '18 at 15:23
atrenet
61
61
add a comment |
add a comment |
1 Answer
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The key fact here is that $f$ is proper and injective. In this case, one can check that $f: M to N$ is an embedding as topological space. So now $f(M)$ is homeomorphic to $M$ and has a natural smooth structure: given a chart $U$ for $M$ , you declare $f(U)$ to be a chart for $f(M)$.
Now ,if you check the definition, this smooth structure implies that $f:M to f(M)$ is smooth and in fact a diffeomorphism: it is a bijective open map being an homeomorphism as you correctly stated, so there is a continuous inverse $g:f(M) to M$. The differential being invertible for every $x in M$ (as you can check from the definition), there exists a local smooth inverse near $f(x)$, which must coincide with the global continous inverse $g$ , as $f$ is bijective.
Now, here it comes the immersion hypothesis. I don't know what is your definition of submanifold. The one I have in mind (which you can check it is equivalent to being the embedded(meaning immersion+topological embedding) image of a manifold) ,is that there are charts $(U,phi)$ on $N$ such that $phi(U cap f(M))=L subset mathbb{R}^n$ ,with $L$ a vector subspace. As you can see, the immersion normal form, gives you this kind of chart for $f(M)$ and so you are done.
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
add a comment |
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The key fact here is that $f$ is proper and injective. In this case, one can check that $f: M to N$ is an embedding as topological space. So now $f(M)$ is homeomorphic to $M$ and has a natural smooth structure: given a chart $U$ for $M$ , you declare $f(U)$ to be a chart for $f(M)$.
Now ,if you check the definition, this smooth structure implies that $f:M to f(M)$ is smooth and in fact a diffeomorphism: it is a bijective open map being an homeomorphism as you correctly stated, so there is a continuous inverse $g:f(M) to M$. The differential being invertible for every $x in M$ (as you can check from the definition), there exists a local smooth inverse near $f(x)$, which must coincide with the global continous inverse $g$ , as $f$ is bijective.
Now, here it comes the immersion hypothesis. I don't know what is your definition of submanifold. The one I have in mind (which you can check it is equivalent to being the embedded(meaning immersion+topological embedding) image of a manifold) ,is that there are charts $(U,phi)$ on $N$ such that $phi(U cap f(M))=L subset mathbb{R}^n$ ,with $L$ a vector subspace. As you can see, the immersion normal form, gives you this kind of chart for $f(M)$ and so you are done.
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
add a comment |
The key fact here is that $f$ is proper and injective. In this case, one can check that $f: M to N$ is an embedding as topological space. So now $f(M)$ is homeomorphic to $M$ and has a natural smooth structure: given a chart $U$ for $M$ , you declare $f(U)$ to be a chart for $f(M)$.
Now ,if you check the definition, this smooth structure implies that $f:M to f(M)$ is smooth and in fact a diffeomorphism: it is a bijective open map being an homeomorphism as you correctly stated, so there is a continuous inverse $g:f(M) to M$. The differential being invertible for every $x in M$ (as you can check from the definition), there exists a local smooth inverse near $f(x)$, which must coincide with the global continous inverse $g$ , as $f$ is bijective.
Now, here it comes the immersion hypothesis. I don't know what is your definition of submanifold. The one I have in mind (which you can check it is equivalent to being the embedded(meaning immersion+topological embedding) image of a manifold) ,is that there are charts $(U,phi)$ on $N$ such that $phi(U cap f(M))=L subset mathbb{R}^n$ ,with $L$ a vector subspace. As you can see, the immersion normal form, gives you this kind of chart for $f(M)$ and so you are done.
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
add a comment |
The key fact here is that $f$ is proper and injective. In this case, one can check that $f: M to N$ is an embedding as topological space. So now $f(M)$ is homeomorphic to $M$ and has a natural smooth structure: given a chart $U$ for $M$ , you declare $f(U)$ to be a chart for $f(M)$.
Now ,if you check the definition, this smooth structure implies that $f:M to f(M)$ is smooth and in fact a diffeomorphism: it is a bijective open map being an homeomorphism as you correctly stated, so there is a continuous inverse $g:f(M) to M$. The differential being invertible for every $x in M$ (as you can check from the definition), there exists a local smooth inverse near $f(x)$, which must coincide with the global continous inverse $g$ , as $f$ is bijective.
Now, here it comes the immersion hypothesis. I don't know what is your definition of submanifold. The one I have in mind (which you can check it is equivalent to being the embedded(meaning immersion+topological embedding) image of a manifold) ,is that there are charts $(U,phi)$ on $N$ such that $phi(U cap f(M))=L subset mathbb{R}^n$ ,with $L$ a vector subspace. As you can see, the immersion normal form, gives you this kind of chart for $f(M)$ and so you are done.
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
The key fact here is that $f$ is proper and injective. In this case, one can check that $f: M to N$ is an embedding as topological space. So now $f(M)$ is homeomorphic to $M$ and has a natural smooth structure: given a chart $U$ for $M$ , you declare $f(U)$ to be a chart for $f(M)$.
Now ,if you check the definition, this smooth structure implies that $f:M to f(M)$ is smooth and in fact a diffeomorphism: it is a bijective open map being an homeomorphism as you correctly stated, so there is a continuous inverse $g:f(M) to M$. The differential being invertible for every $x in M$ (as you can check from the definition), there exists a local smooth inverse near $f(x)$, which must coincide with the global continous inverse $g$ , as $f$ is bijective.
Now, here it comes the immersion hypothesis. I don't know what is your definition of submanifold. The one I have in mind (which you can check it is equivalent to being the embedded(meaning immersion+topological embedding) image of a manifold) ,is that there are charts $(U,phi)$ on $N$ such that $phi(U cap f(M))=L subset mathbb{R}^n$ ,with $L$ a vector subspace. As you can see, the immersion normal form, gives you this kind of chart for $f(M)$ and so you are done.
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
answered Dec 13 '18 at 7:47
Tommaso Scognamiglio
453312
453312
add a comment |
add a comment |
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