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I am trying to understand the concept of embedded submanifolds and have the following understanding:

Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $m\gt n$ .Let $F\colon N\to M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $F\colon N \to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $f\circ F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.

Now, the definition of embedded submanifolds as given in the text of boothby is:

image of a topological embedding+immersion is an embedded submanifold

My problem is:

If just taking $F\colon N \to F(N)$ as a topological embedding gives $F(N)$ diffeomorphic to $N$. So, it gives a smooth manifold $F(N)$ with same dimension as $N$, why do we even need to consider immersions?

Thanks in advance

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  • $\begingroup$ What is the definition of topological embedding here? $\endgroup$ – Anubhav Mukherjee Jan 10 at 18:48
  • $\begingroup$ @Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology $\endgroup$ – Abhishek Shrivastava Jan 10 at 18:50
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Let $N=\Bbb R$ and $M=\Bbb R^2$. The map $F\colon N\to M$, $x\mapsto (x^3,0)$ is a topological embedding of $\Bbb R$ into $\Bbb R^2$. However, the image $F(N)=\Bbb R\times 0\subset \Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.) To avoid this, you want $F$ to be an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same.


From $M$ the subspace $\Bbb R\times 0$ inherits a smooth structure with atlas defined by the global chart $(x,0)\mapsto x$. Pushed forward via $F$, we get a smooth structure on $\Bbb R\times 0$ with atlas defined by the global chart $(x,0)\mapsto x^{1/3}$. Now check that with respect to these two atlases the identity map $\Bbb R\times 0 \to \Bbb R\times 0$ is not a diffeomorphism, so the two smooth structures are different. They still define diffeomorphic manifolds, since $(x,0)\mapsto (x^3,0)$ is a diffeomorphism with respect to these two atlases.

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  • $\begingroup$ :thanks for the answer.could you elaborate:"However, the image $F(N)=\Bbb R\times 0\subset \Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$" $\endgroup$ – Abhishek Shrivastava Jan 10 at 19:05
  • $\begingroup$ I added some details, does this help? $\endgroup$ – Christoph Jan 10 at 19:14
  • $\begingroup$ :understood....thanks a lot $\endgroup$ – Abhishek Shrivastava Jan 10 at 19:16
  • $\begingroup$ :just to ask one more thing: $\endgroup$ – Abhishek Shrivastava Jan 10 at 19:16
  • $\begingroup$ :just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $\pmb N$ with a n dimensional subspace of tangent space at f(p) of $\pmb M$"is this correct?? $\endgroup$ – Abhishek Shrivastava Jan 10 at 19:25

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