If $\phi:M\longrightarrow N$ is an injective smooth map between two manifolds, then is $d\phi_m:M_m\longrightarrow N_{\phi(m)}$, the induced map between the tangent spaces injective too?
I tried the following : If $v\in M_m$ is such that $d\phi_m(v)=0$, then for all $g$, $C^{\infty}$ function in a neighbourhood of $\phi(m)$, $d\phi_m(v)(g)=0$, that is $v(g\circ\phi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.
I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $\phi$ being injective will automatically make it an immersion.
Best Answer
The answer is no. Let $\phi:\mathbb R\rightarrow\mathbb R^2$ be $\phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.