Injectivity of the derivative of an embedding

Question

I want to prove the following:

If $i:M^{m}\rightarrow N^{n} $ is an embedding (i.e. it is a diffeomorphism onto its image, and its image is a differential manifold), then $i_{*}(p):T_{p}M\rightarrow T_{i(p)}N $ is an injective (linear) map (for any $p\in M $).

My attempt: If I prove that the rank of $i_{*}(p) $ is $m$, then, since $$\mbox{dim}(T_{p}M)=\mbox{dim}(\mbox{ker}(i_{*}(p))+\mbox{dim}(i_{*}(p)(T_{p}M))=\mbox{dim}(\mbox{ker}(i_{*}(p))+\mbox{rank}(i_{*}(p) $$ and $\mbox{dim}(T_{p}M)=m $, then $\mbox{dim}(\mbox{ker}(i_{*}(p))=0 $ and so $i_{*}(p) $ is injective. So, how can I prove that the rank of $i_{*}(p) $ is $m $?

Also, my teacher said something about trying to use “local coordinates where $i:\mathbb{R}^{m}\hookrightarrow\mathbb{R}^{n} $” to solve this. If you have a clue of what this means, please tell me.

Answer 1

Any immersion looks locally like the "canonical immersion" $i\colon\Bbb R^m\to\Bbb R^n$, given by $i(x_1,\dots,x_m) = (x_1,\dots,x_m,0,\dots,0)$. The derivative map is the inclusion map.

But you can argue this more directly (and slightly less generally). If $f\colon M\to f(M)\subset N$ is a diffeomorphism to its image, let $g$ be the (smooth) inverse function. Since $g\circ f = \text{id}_M$, we have $dg_{f(x)}\circ df_x = \text{id}_{T_xM}$, and so by easy set theory $df_x$ is injective as a map $T_xM\to T_{f(x)}N$.