Question

Suppose $M$ and $N$ are smooth manifolds. An immersion is a smooth map $f: M \rightarrow N$ whose pushforward is injective at each point.

Is a smooth injective map an immersion?

We can actually simplify the question further.

Suppose $f : M \rightarrow N$ is a smooth injective map. Suppose $(U, \phi)$ and $(V, \psi)$ are smooth charts for $M$ and $N$ respectively. Fix $p \in U$. Then

$$ f_\ast = ( \psi^{-1}\circ \psi \circ f \circ \phi^{-1} \circ \phi)_{\ast} = (\psi^{-1})_\ast \circ (\psi \circ f \circ \phi^{-1})_\ast \circ \phi_\ast $$

As $\phi$ and $\psi$ are diffeomorphisms, $\phi_\ast$ and $(\psi^{-1})_\ast$ are linear isomorphisms.

Therefore, if $(\psi \circ f \circ \phi^{-1})_\ast$ is injective then $f_\ast$ is injective.

This shows that if every smooth injective map between open subsets of euclidean space is an immersion, then every smooth injective map between smooth manifolds is an immersion.

Answer 1

The answer is no.

Here's an example --- which is based on an example I gave in my multivariable calculus discussion section recently!! :-)

Let M be (0,pi) with coordinate t. Let N be R^2 with coordinates x and y. Define a map x(t)=sin(t), y(t)=sin(t) for t in (0,pi/2] and x(t)=2-cos(t-pi/2), y(t)=2-cos(t-pi/2) for t in (pi/2,pi). This map is smooth and injective on points but the derivative is zero at t=pi/2.

Edit: Ok, I got a bit too excited about using multivariable calculus. As algori mentions, t^3 is a simpler example. Or just taking one of the coordinate functions in my original example works too. There's no need for multivariable calculus.

Answer 2

For a smooth injective map $f:M\rightarrow N$, there is only an obstruction to be $f$ an immersion, and it is that its rank is not constant.