# Is a smooth injective map an immersion?

I am trying to work out some basic properties of immersions/submersions/embeddings etc. Is the following reasoning correct?
Let $M,N$ be smooth manifolds, and $f:M \rightarrow N$ a smooth injective map. Since $rk (f_*)$ is equal to the dimension of $M$, then $f$ is a map of constant rank, and hence an embedding.

The conclusion $rk(f_*)$ is equal to the dimension of $M$ is not correct. Consider the map $f : \mathbb{R} \to \mathbb{R}$ given by $f(x)=x^3$. The rank of its derivative at $x=0$ is 0.

Consider a map $f : M \to N$. Let $f_*$ denote the derivative of $f$. Let's recall some definitions:

• Immersion: $f_*$ is everywhere injective

• Embedding: $f_*$ is everywhere injective + $f$ is a homeomorphism onto its image.

Example of immersion that is not an embedding: immersing the circle as a figure eight on the plane $\gamma : (-\frac{\pi}{2},\frac{3\pi}{2}) \to \mathbb{R}^2$ given by $\gamma(t)=(\sin(2t),\cos(t))$. It is injective. But it is not an embedding: the figure eight is not compact in the subspace topology of $\mathbb{R}^2$, whereas the domain is compact in the subspace topology coming from $\mathbb{R}$.

On the other hand, if a smooth injective map has constant rank then it is an immersion. This follows from the constant rank theorem, i.e. that in local coordinates any smooth map $f : M \to N$, $m=\dim M$, $n=\dim N$, with constant rank $k$ can be written as $(x_1,\cdots,x_k,x_{k+1},\cdots, x_m) \mapsto (x_1,\cdots,x_k,0,\cdots,0)$. Finally, if $f: M \to N$ is an injective immersion and either $M$ is compact or $f$ is proper, then $f$ is an embedding.

• I see, thanks. What if I also demand that $f$ has a smooth inverse (i.e. f is a diffeomorphism between $M$ and its image)? – Orpheus Feb 11 '18 at 11:20
• If $f$ is a diffeomorphism then $f$ is, thus, a homeomorphism, so $f : M \to N$ is a homeomorphism onto its image. To see that $f_*$ is everywhere injective, note that since $f$ is a diffeomorphism, $\operatorname{id}=f \circ f^{-1}$. So, taking local coordinates $(x_1,\cdots,x_n) \in U \subset \mathbb{R}^n$ about a point $x \in M$, and using chain rule: $\operatorname{Id}_{\mathbb{R}^n} = f_*^{-1}(f(x_1,\cdots,x_n))f_*(x_1,\cdots,x_n)$. This means that $f_*(x_1,\cdots,x_n)$ must have full-rank at every $x \in M$ (since $\operatorname{Id}_{\mathbb{R}^n}$ has obviously constant full-rank). – DMG Feb 11 '18 at 15:06
• Got it, thanks a lot. – Orpheus Feb 11 '18 at 15:33