Let $U$ be an open subset of $\mathbb R^n$, $f:U\rightarrow\mathbb R^k$ a smooth map such that its pushforward is onto, for each $x\in U$, i.e. $$f_{*x}:T_xU\rightarrow T_{f(x)}\mathbb R^k$$ is onto , for each $x\in U$. We set $M=f^{-1}(0)$. Then prove that $$T_xM=\ker f_{*x}$$
To prove this, first of all, from a known theorem we see that $M$ is a smooth manifold of dimension $\dim M=n-k$, and it is also a submanifold of $\mathbb R^n$. Therefore, we can consider the inclusion map:
$$i:M\rightarrow \mathbb R^n$$ $$i(x^1,x^2,\dots,x^{n-k})=(x^1,x^2,\dots,x^{n-k},0,\dots,0)$$ which is an embedding. Take the pushforward:$$i_*:TM\rightarrow T\mathbb R^n$$ Then for $x\in M$, $i_{*x}:T_xM\rightarrow T_{i(x)}\mathbb R^n=\mathbb R^n$
Can you help me how to proceed?
What is the dimension of $\ker f_{*x}$? Can you show $T_x M\subset\ker f_{*x}$?