# Images of immersed submanifolds are immersed submanifolds as well.

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.
• But $i(N)$ is given a manifold structure by $N$. – user555729 May 1 '19 at 13:49
• @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. – Adam Chalumeau May 1 '19 at 14:06