# Understanding Embedded submanifolds and immersions

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?

• What is the definition of topological embedding here? – Anubhav Mukherjee Jan 10 at 18:48
• @Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology – Abhishek Shrivastava Jan 10 at 18:50

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.
• :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 $N$" – Abhishek Shrivastava Jan 10 at 19:05
• :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?? – Abhishek Shrivastava Jan 10 at 19:25