Let $f: M \to N$ be a smooth embedding of differentiable manifolds of dimensions $m,n$ respectively. Let $L \subset M$ be a smooth submanifold of dimensions $1 \leq k \leq m$. Is it true that $df$, restricted to $L$, will have rank $k$, i.e. $L$ is also smoothly embedded in $N$ by $f$?
My approach goes along the lines that given a point $p \in L$, the differential at $p$ has as domain the $k$ dimensional vector space $TL_p$, which is a subspace of $TM_p$. So restricting $df$ to $TL_p$ will still give an injective map, which must have rank $k$. Does this work? Am I missing something?
If $L$ is an embedded submanifold, then $f:L\to N$ is smooth embedding since the inclusion $i:L\to M$ is smooth embedding and composition of smooth embeddings is smooth embedding.
But if $L$ is only an immersion submanifold, it could happen that $f_{|L}$ is not smooth embedding. Take the classical example: let $\beta:(-\pi,\pi)\to\mathbb{R}, \beta(t)=(\sin 2t,\sin t)$, $L$ be the imagem of $\beta$ with topology given by $\beta$ (declare that $\beta$ is a homeomorphism) and take $M=N=\mathbb{R}^2, f=\mathrm{id}_{\mathbb{R}^2}$. Now, let $L'$ be the imagem of $\beta$ with subspace topology, then $L'$ is a compact set but $L$ is not ($L$ is homeomorphic to an open interval), thus $f:L\to f(L)=L'$(in the image we have to consider the subspace topology, by definition) is not a homeomorphism.