Is an injective smooth map an immersion? - MathOverflow

Is an injective smooth map an immersion?

2010-02-15T05:36:36Z

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 by Kevin Lin for Is an injective smooth map an immersion?

2010-02-15T05:55:36Z

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 by Giuseppe Tortorella for Is an injective smooth map an immersion?

2011-02-06T15:38:30Z

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.