I Injective immersion that is not a smooth embedding

cianfa72

Hi, I'm aware of a typical example of injective immersion that is not a topological embedding: figure 8

$\beta: (-\pi, \pi) \to \mathbb R^2$, with $\beta(t)=(\sin 2t,\sin t)$

As explained here an-injective-immersion-that-is-not-a-topological-embedding the image of $\beta$ is compact in $\mathbb R^2$ subspace topology while the domain open interval is not, thus $\beta$ is not a smooth embedding.

Consider it from the point of view of "homeomorphism onto its image" definition, I was trying to find out an instance of image subset open in the subspace topology that actually is not open in the domain topology or the other way around.

Can you help me ? Thanks.

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WWGD

Gold Member
Hi, I'm aware of a typical example of injective immersion that is not a topological embedding: figure 8

$\beta: (-\pi, \pi) \to \mathbb R^2$, with $\beta(t)=(\sin 2t,\sin t)$

As explained here an-injective-immersion-that-is-not-a-topological-embedding the image of $\beta$ is compact in $\mathbb R^2$ subspace topology while the domain open interval is not, thus $\beta$ is not a smooth embedding.

Consider it from the point of view of "homeomorphism onto its image" definition, I was trying to find out an instance of image subset open in the subspace topology that actually is not open in the domain topology or the other way around.

Can you help me ? Thanks.
You can show the complement in the image is open, to show the image itself is closed. Take any point x in the complement. It is at a non-zero distance from the image. Consider a ball centered at x with distance d/2 from the image, to show complement is open, so image itself is closed.

cianfa72

Thinking again about it, I believe we can find an example of that subset looking at the inverse map $\beta^{-1}$.
Take an open subset $(-\varepsilon, \varepsilon)$ belonging to the $\mathbb R^1$ subspace topology induced on $(-\pi,\pi)$. Its preimage under $\beta^{-1}$ includes the Figure 8 central point and is not an open subset of Figure 8 endowed with $\mathbb R^2$ subspace topology (it cannot be obtained through the intersection of an $\mathbb R^2$ open subset with the Figure 8 itself) therefore $\beta$ is not an homomorphism with the image (in the subspace topology)

Is that right ?

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cianfa72

Thinking again about it, I believe we can find an example of that subset looking at the inverse map $\beta^{-1}$.
Take an open subset $(-\varepsilon, \varepsilon)$ belonging to the $\mathbb R^1$ subspace topology induced on $(-\pi,\pi)$. Its preimage under $\beta^{-1}$ includes the Figure 8 central point and is not an open subset of Figure 8 endowed with $\mathbb R^2$ subspace topology (it cannot be obtained through the intersection of an $\mathbb R^2$ open subset with the Figure 8 itself) therefore $\beta$ is not an homomorphism with the image (in the subspace topology)

Is that right ?
Help ! I'm a beginner...just to check I got it correctly, can you confirm that ? Thanks

WWGD

Gold Member
Thinking again about it, I believe we can find an example of that subset looking at the inverse map $\beta^{-1}$.
Take an open subset $(-\varepsilon, \varepsilon)$ belonging to the $\mathbb R^1$ subspace topology induced on $(-\pi,\pi)$. Its preimage under $\beta^{-1}$ includes the Figure 8 central point and is not an open subset of Figure 8 endowed with $\mathbb R^2$ subspace topology (it cannot be obtained through the intersection of an $\mathbb R^2$ open subset with the Figure 8 itself) therefore $\beta$ is not an homomorphism with the image (in the subspace topology)

Is that right ?
Careful, homeomorphism, not homomorphism. But be careful, a map may be open --taking open sets to open sets -- and not be a homeomorphism.

WWGD

Gold Member
Thinking again about it, I believe we can find an example of that subset looking at the inverse map $\beta^{-1}$.
Take an open subset $(-\varepsilon, \varepsilon)$ belonging to the $\mathbb R^1$ subspace topology induced on $(-\pi,\pi)$. Its preimage under $\beta^{-1}$ includes the Figure 8 central point and is not an open subset of Figure 8 endowed with $\mathbb R^2$ subspace topology (it cannot be obtained through the intersection of an $\mathbb R^2$ open subset with the Figure 8 itself) therefore $\beta$ is not an homomorphism with the image (in the subspace topology)

Is that right ?
Just remember that a map may be open but not a homeomorphism. I think this map is a local diffeomorphism, so it sends open to open. .

cianfa72

Just remember that a map may be open but not a homeomorphism. I think this map is a local diffeomorphism, so it sends open to open. .
ok, but just for the very fact exists an open set that $\beta$ does not send in an open set, it suffices to say it is not homeomorphism, do you ?

WWGD

Gold Member
ok, but just for the very fact exists an open set that $\beta$ does not send in an open set, it suffices to say it is not homeomorphism, do you ?
, Yes, that is enough to guarantee a maps not a diffeomorphism

EDIT: But as you said, there is an open interval in #(-\epsilon, \epsilon)# that hits the center part of the figure 8, whose image is not open.

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