# Image of an Immersion

I cannot prove any conclusion of this problem. Can anyone please help me? Let $f:M\to N$ be an immersion of $M$ into $N$ and dim $M=\dim N$.Prove or disprove that $f(M)$ is a submanifold. Thanks for any help.

• The image $f(M)$ need not be a submanifold as there might be self-intersections. Consider Boy's surface for example en.wikipedia.org/wiki/Boy%27s_surface (which gives an immersion of the real projective plane into $\mathbb{R}^3$), or something even less exotic such as the surface parametrized by $$\mathbf{x}(u, v) = \left(\cos\left(u\right), \sin\left(2 u\right), v\right),$$ where $-\pi < u < \pi$ and $0 < v < 3$. (I would add a figure for the second example, but comments don't seem to allow for that). If you think more locally in $M$, then you can draw some conclusions. – THW Mar 8 '14 at 15:12
• math.stackexchange.com/questions/568160/… .Have a look at this.By this result I think it will be a submanifold,because in this case we have a local diffeomorphism. – Ester Mar 8 '14 at 16:09
• I missed the assumption in the question that the dimension of $M$ was equal to the dimension of $N$ and this certainly changes the relevancy of my comment above. – THW Mar 8 '14 at 16:30

$f(M)$ is an open subset of $N$ and thus a submanifold. Proof by unwrapping of definitions: say $y=f(x)\in f(M)$. Since $f$ is an immersion $df(x)$ is injective, and since $\dim(M)=\dim(N)$ $df(x)$ is in fact an isomorphism. By the inverse function theorem $f$ restricted to some neighborhood of $x$ is a diffeomorphism onto a neighborhood of $y$. In particular, $f(M)$ contains a neighborhood of $y$.
1. Use this. An immersion is an open map if (and only if, I think) $$\dim M = \dim N$$. This is because for $$\dim M = \dim N$$, we have that immersion, submersion are local diffeomorphism are equivalent.