Mixing
Homeomorphism wiht image and diffeomorphism with image
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I was thinking of sufficient conditions for a map $F:Xto Y_1times Y_2$ to be homeomorphism with its image. So I thought the following facts and proofs:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are continuous and $F_1$ is an homemorphism with its image, then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is an homemoprhism with its image.
Proof.
$overlineF:Xto F(X)$ si clearly continuous and bijective. Its inverse is $overlineF^-1:F(X)to X, quad (x,y)mapsto F_1^-1(x)$ wich is continuous since $overlineF^-1=
overlineF_1^-1 circ (pr_1)_F(X):F(X)to F_1(X)to X$.
Then i was thinking if I could "generalize" this to smooth maps.
So i thought:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are $C^infty$ and $F_1$ is a differomorpshism with its image (assuming $F_1(X)$ embedded submanifold of $Y_1$), then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is a diffeomorphism with its image (assuming that $F(X)$ is an embedded submanifold of $Y_1times Y_2$).
Proof.
Since $F_1$ and $F_2$ are $C^infty$ then $F$ is $C^infty$. Since $F(X)$ is an embedded submanifold of $Y_1times Y_2$ then also $overlineF:Xto F(X)$ is $C^infty$. The map $pr_1:Y_1 times Y_2to Y_1$ is $C^infty$, so also $pr_1:F(X)to Y_1$ is $C^infty$ so also $(pr_1)_F(X):F(X)to F_1(X)$ is $C^infty$. The map $overlineF_1:Xto F_1(X)$ is a diffeomorphism. Then $overlineF^-1=overlineF_1^-1 circ (pr_1)_F(X):F(X)to X$ is $C^infty$.
My questions:
1) Are these facts and proofs correct?
2) Are there other (frequently used and "useful") sufficient conditions for a map to be an heoomorphism with its image or differomorpshims with its image?
3) Are the hypothesis: $F(X)$ embedded submanifold of $Y_1times Y_2$ and $F_1(X)$ embedded submanifold of $Y_1$ necesary?
general-topology proof-verification smooth-manifolds
asked Jul 22 at 18:40
Minato
184111
add a comment |Â
up vote
1
down vote
favorite
I was thinking of sufficient conditions for a map $F:Xto Y_1times Y_2$ to be homeomorphism with its image. So I thought the following facts and proofs:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are continuous and $F_1$ is an homemorphism with its image, then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is an homemoprhism with its image.
Proof.
$overlineF:Xto F(X)$ si clearly continuous and bijective. Its inverse is $overlineF^-1:F(X)to X, quad (x,y)mapsto F_1^-1(x)$ wich is continuous since $overlineF^-1=
overlineF_1^-1 circ (pr_1)_F(X):F(X)to F_1(X)to X$.
Then i was thinking if I could "generalize" this to smooth maps.
So i thought:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are $C^infty$ and $F_1$ is a differomorpshism with its image (assuming $F_1(X)$ embedded submanifold of $Y_1$), then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is a diffeomorphism with its image (assuming that $F(X)$ is an embedded submanifold of $Y_1times Y_2$).
Proof.
Since $F_1$ and $F_2$ are $C^infty$ then $F$ is $C^infty$. Since $F(X)$ is an embedded submanifold of $Y_1times Y_2$ then also $overlineF:Xto F(X)$ is $C^infty$. The map $pr_1:Y_1 times Y_2to Y_1$ is $C^infty$, so also $pr_1:F(X)to Y_1$ is $C^infty$ so also $(pr_1)_F(X):F(X)to F_1(X)$ is $C^infty$. The map $overlineF_1:Xto F_1(X)$ is a diffeomorphism. Then $overlineF^-1=overlineF_1^-1 circ (pr_1)_F(X):F(X)to X$ is $C^infty$.
My questions:
1) Are these facts and proofs correct?
2) Are there other (frequently used and "useful") sufficient conditions for a map to be an heoomorphism with its image or differomorpshims with its image?
3) Are the hypothesis: $F(X)$ embedded submanifold of $Y_1times Y_2$ and $F_1(X)$ embedded submanifold of $Y_1$ necesary?
general-topology proof-verification smooth-manifolds
asked Jul 22 at 18:40
Minato
184111
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was thinking of sufficient conditions for a map $F:Xto Y_1times Y_2$ to be homeomorphism with its image. So I thought the following facts and proofs:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are continuous and $F_1$ is an homemorphism with its image, then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is an homemoprhism with its image.
Proof.
$overlineF:Xto F(X)$ si clearly continuous and bijective. Its inverse is $overlineF^-1:F(X)to X, quad (x,y)mapsto F_1^-1(x)$ wich is continuous since $overlineF^-1=
overlineF_1^-1 circ (pr_1)_F(X):F(X)to F_1(X)to X$.
Then i was thinking if I could "generalize" this to smooth maps.
So i thought:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are $C^infty$ and $F_1$ is a differomorpshism with its image (assuming $F_1(X)$ embedded submanifold of $Y_1$), then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is a diffeomorphism with its image (assuming that $F(X)$ is an embedded submanifold of $Y_1times Y_2$).
Proof.
Since $F_1$ and $F_2$ are $C^infty$ then $F$ is $C^infty$. Since $F(X)$ is an embedded submanifold of $Y_1times Y_2$ then also $overlineF:Xto F(X)$ is $C^infty$. The map $pr_1:Y_1 times Y_2to Y_1$ is $C^infty$, so also $pr_1:F(X)to Y_1$ is $C^infty$ so also $(pr_1)_F(X):F(X)to F_1(X)$ is $C^infty$. The map $overlineF_1:Xto F_1(X)$ is a diffeomorphism. Then $overlineF^-1=overlineF_1^-1 circ (pr_1)_F(X):F(X)to X$ is $C^infty$.
My questions:
1) Are these facts and proofs correct?
2) Are there other (frequently used and "useful") sufficient conditions for a map to be an heoomorphism with its image or differomorpshims with its image?
3) Are the hypothesis: $F(X)$ embedded submanifold of $Y_1times Y_2$ and $F_1(X)$ embedded submanifold of $Y_1$ necesary?
general-topology proof-verification smooth-manifolds
asked Jul 22 at 18:40
Minato
184111
I was thinking of sufficient conditions for a map $F:Xto Y_1times Y_2$ to be homeomorphism with its image. So I thought the following facts and proofs:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are continuous and $F_1$ is an homemorphism with its image, then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is an homemoprhism with its image.
Proof.
$overlineF:Xto F(X)$ si clearly continuous and bijective. Its inverse is $overlineF^-1:F(X)to X, quad (x,y)mapsto F_1^-1(x)$ wich is continuous since $overlineF^-1=
overlineF_1^-1 circ (pr_1)_F(X):F(X)to F_1(X)to X$.
Then i was thinking if I could "generalize" this to smooth maps.
So i thought:
If $F_1:Xto Y_1$ and $F_2: X to Y_2$ are $C^infty$ and $F_1$ is a differomorpshism with its image (assuming $F_1(X)$ embedded submanifold of $Y_1$), then $F:Xto Y_1times Y_2, quad xmapsto(F_1(x), F_2(x))$ is a diffeomorphism with its image (assuming that $F(X)$ is an embedded submanifold of $Y_1times Y_2$).
Proof.
Since $F_1$ and $F_2$ are $C^infty$ then $F$ is $C^infty$. Since $F(X)$ is an embedded submanifold of $Y_1times Y_2$ then also $overlineF:Xto F(X)$ is $C^infty$. The map $pr_1:Y_1 times Y_2to Y_1$ is $C^infty$, so also $pr_1:F(X)to Y_1$ is $C^infty$ so also $(pr_1)_F(X):F(X)to F_1(X)$ is $C^infty$. The map $overlineF_1:Xto F_1(X)$ is a diffeomorphism. Then $overlineF^-1=overlineF_1^-1 circ (pr_1)_F(X):F(X)to X$ is $C^infty$.
My questions:
1) Are these facts and proofs correct?
2) Are there other (frequently used and "useful") sufficient conditions for a map to be an heoomorphism with its image or differomorpshims with its image?
3) Are the hypothesis: $F(X)$ embedded submanifold of $Y_1times Y_2$ and $F_1(X)$ embedded submanifold of $Y_1$ necesary?
general-topology proof-verification smooth-manifolds
asked Jul 22 at 18:40
Minato
184111
asked Jul 22 at 18:40
Minato
184111
asked Jul 22 at 18:40
Minato
184111
asked Jul 22 at 18:40
Minato
184111
184111
add a comment |Â
add a comment |Â
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The first fact and proof are correct. The second is as well. More statements concerning the matter are as follows:
1) Many categories admit products. In this case, morphisms $F_1: X to Y_1$ and $F_2: X to Y_2$, where $X, Y_1, Y_2$ are objects of the respective categories, automatically yield a morphism of that category into the product $Y_1 times Y_2$. Since the projections are also maps in the respective category, the partial inverse argument of yours works in this generality.
2) What immediately comes to mind is the inverse function theorem, which given injectivity and a nonsingular differential yields exactly what you want. Note that the inverse function theorem also holds on manifolds.
3) That depends what kind of a morphism you want. If you want a morphism of manifolds, then certainly you have to make sure that the domain of definition is also a manifold. Otherwise, you may need to restrict to categories which allow more "unorthodox" subobjects; for instance, the category of topological spaces.
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The first fact and proof are correct. The second is as well. More statements concerning the matter are as follows:
1) Many categories admit products. In this case, morphisms $F_1: X to Y_1$ and $F_2: X to Y_2$, where $X, Y_1, Y_2$ are objects of the respective categories, automatically yield a morphism of that category into the product $Y_1 times Y_2$. Since the projections are also maps in the respective category, the partial inverse argument of yours works in this generality.
2) What immediately comes to mind is the inverse function theorem, which given injectivity and a nonsingular differential yields exactly what you want. Note that the inverse function theorem also holds on manifolds.
3) That depends what kind of a morphism you want. If you want a morphism of manifolds, then certainly you have to make sure that the domain of definition is also a manifold. Otherwise, you may need to restrict to categories which allow more "unorthodox" subobjects; for instance, the category of topological spaces.
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
add a comment |Â
up vote
0
down vote
The first fact and proof are correct. The second is as well. More statements concerning the matter are as follows:
1) Many categories admit products. In this case, morphisms $F_1: X to Y_1$ and $F_2: X to Y_2$, where $X, Y_1, Y_2$ are objects of the respective categories, automatically yield a morphism of that category into the product $Y_1 times Y_2$. Since the projections are also maps in the respective category, the partial inverse argument of yours works in this generality.
2) What immediately comes to mind is the inverse function theorem, which given injectivity and a nonsingular differential yields exactly what you want. Note that the inverse function theorem also holds on manifolds.
3) That depends what kind of a morphism you want. If you want a morphism of manifolds, then certainly you have to make sure that the domain of definition is also a manifold. Otherwise, you may need to restrict to categories which allow more "unorthodox" subobjects; for instance, the category of topological spaces.
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The first fact and proof are correct. The second is as well. More statements concerning the matter are as follows:
1) Many categories admit products. In this case, morphisms $F_1: X to Y_1$ and $F_2: X to Y_2$, where $X, Y_1, Y_2$ are objects of the respective categories, automatically yield a morphism of that category into the product $Y_1 times Y_2$. Since the projections are also maps in the respective category, the partial inverse argument of yours works in this generality.
2) What immediately comes to mind is the inverse function theorem, which given injectivity and a nonsingular differential yields exactly what you want. Note that the inverse function theorem also holds on manifolds.
3) That depends what kind of a morphism you want. If you want a morphism of manifolds, then certainly you have to make sure that the domain of definition is also a manifold. Otherwise, you may need to restrict to categories which allow more "unorthodox" subobjects; for instance, the category of topological spaces.
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
The first fact and proof are correct. The second is as well. More statements concerning the matter are as follows:
1) Many categories admit products. In this case, morphisms $F_1: X to Y_1$ and $F_2: X to Y_2$, where $X, Y_1, Y_2$ are objects of the respective categories, automatically yield a morphism of that category into the product $Y_1 times Y_2$. Since the projections are also maps in the respective category, the partial inverse argument of yours works in this generality.
2) What immediately comes to mind is the inverse function theorem, which given injectivity and a nonsingular differential yields exactly what you want. Note that the inverse function theorem also holds on manifolds.
3) That depends what kind of a morphism you want. If you want a morphism of manifolds, then certainly you have to make sure that the domain of definition is also a manifold. Otherwise, you may need to restrict to categories which allow more "unorthodox" subobjects; for instance, the category of topological spaces.
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
answered Jul 22 at 18:58
AlgebraicsAnonymous
69111
69111
add a comment |Â
add a comment |Â
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