show/hide this revision's text 3 added 245 characters in body

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.

show/hide this revision's text 2 added 2 characters in body

The answer is no.

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

Let X M be (0,pi) with coordinate t. Let Y N be R^2 with coordinates x and y. Define a map x(t)=sin tx(t)=sin(t), y(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.
show/hide this revision's text 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 X be (0,pi) with coordinate t. Let Y 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.