Let $i:N\to M$ be an injective immersion. $(N,i)$ is called an immersed submanifold of $M$. In this case, $i(N)$ is given a manifold structure from the smooth structure of $N$.

Now consider the inclusion map $i: i(N)\to M$. Is it true that $(i(N),i)$ is an immersed submanifold of $M$?

This question came to my mind when I verified that a nonvanishing integral curve with the inclusion map is an immersed submanifold


To talk about an immersion you need to have a map $j:M_1\to M_2$ where $M_1$ and $M_2$ are manifolds. Here you have a map $i:i(M)\to N$ but $i(M)$ need not be a manifold so it doesn't make sens to be an immersion in this case. So the answer is no.

Edit: I went a bite fast here. Lets call $S=i(M)$ with the differential structure given by $M$. Lets call $j:S\to N$ the map you are looking at. Then $h:M\to S$ defined by $h(m)=i(m)$ is a diffeomorphism (almost by definition). The map you are looking for is just $j=i\circ h^{-1}$, which is an immersion.

  • $\begingroup$ But $i(N)$ is given a manifold structure by $N$. $\endgroup$ – user555729 May 1 '19 at 13:49
  • $\begingroup$ @User12239 I read a bite fast I edited the post. The thing is you should change of notation for $i(M)$ because written like this it looks like it has been given the usual topology. $\endgroup$ – Adam Chalumeau May 1 '19 at 14:06
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    $\begingroup$ Yes so I didn’t need to verify that an integral curve is an immersion using its derivatives as tangents because this is a general fact for all immersions $\endgroup$ – user555729 May 1 '19 at 14:13

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