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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.

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    $\begingroup$ What about $f(x)=x^3$ from $\mathbb R$ to $\mathbb R$..? $\endgroup$
    – diverietti
    Commented Feb 6, 2011 at 18:30

2 Answers 2

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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.

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  • $\begingroup$ To reformulate @argtororella, a smooth injective map of constant rank is an immersion. You can also see wikipedia, constant rank for further details. $\endgroup$
    – Picaud Vincent
    Commented Jun 27, 2018 at 18:28
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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 $\mathbb R^2$ 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.

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    $\begingroup$ I give a homework problem to my manifolds students ,asking them to construct a smooth injection (an immersion at all but one point)from $\mathbb R$ and onto the "corner space" $(x,y)$ such that $x \geq 0$ and $y=0$ union $x=0$ and $y \geq 0$. $\endgroup$
    – Ryan Budney
    Commented Feb 15, 2010 at 6:03
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    $\begingroup$ Kevin -- this is not continuous;) On a more serious level: $t\mapsto t^3$ would do, no multivariate calculus necessary. $\endgroup$
    – algori
    Commented Feb 15, 2010 at 6:03
  • $\begingroup$ algori: I fixed it. And yeah that works too. $\endgroup$
    – Kevin H. Lin
    Commented Feb 15, 2010 at 6:07
  • $\begingroup$ thanks for the help! The cubic example was just what i was looking for. i feel a bit silly now though $\endgroup$
    – Daniel Barter
    Commented Feb 15, 2010 at 6:20
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    $\begingroup$ It's interesting if you think about this in terms of Toen's "Geometric Contexts". I see it as a failure of the classical theory that injective morphisms do not correspond to immersions. Rather, if we replace the classical notions with functors of points, a lot of these "degeneracies", if they could be called that, disappear to some extent. This is because injective morphisms (our choice of morphisms depends on the geometric context, of course) do in fact become "open" immersions. $\endgroup$
    – Harry Gindi
    Commented Feb 15, 2010 at 8:56

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