Return to Answer

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.

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.

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.

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.

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.

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.