Let $V\_1, V\_2 \rightarrow M $ be smooth vector bundles over a manifold $M$ and $s\_1: M \rightarrow V\_1$ 
a smooth section transverse to the zero set and $s\_2: M \rightarrow V\_2$ a continuous section 
such that $ s\_2 : s\_1^{-1}(0) \rightarrow V\_2$ is a smooth section, transverse to the zero 
set. Let $p\in M$ be a point such that 
$$ s\_1(p)=0, \qquad s\_2(p) =0. $$
Does there always exist a solution $p(t\_1, t\_2)$ near $p$ such that 
$$ s\_1(p(t\_1,t\_2)) =t\_1 \qquad s\_2(p(t\_1,t\_2)) = t\_2 $$ 
if $t\_1$ and $t\_2$ are sufficiently small? To make sense of sufficiently small we can 
just chose some metric on $V\_1$ and $V\_2$. 


Note that if I said that the section $p \rightarrow s\_1(p) \oplus s\_2(p)$ is smooth 
and transverse to the zero set, then the answer is obviously yes (by implicit function 
theorem). It seems that one should be able to imitate the proof of implicit function 
theorem (ie contraction mapping principal) to justify what I am asking. But I am not sure 
of this. 



A version of implicit function theorem when sections are not everywhere smooth?
Note that by your assumption $s\_2$ is only smooth along the submanifold $s^{-1}(0)$ of $M$ and might even jump in directions transverse to it. So it is easy to come up with a counterexample: $M=\mathbb R^2$, $s\_1(x,y)=x$, $s\_2(x,y) = y$ for $x\le 0$ and $=y+1$ for $x>0$. 


But if you assume that $s\_2$ is strictly differentiable on $M$ along $s\_1^{-1}(0)$, then is true in a somewhat weaker sense. See 


* MR0817719 Reviewed Ver Eecke, Paul Applications du calcul différentiel. (French) [Applications of differential calculus] Mathématiques. [Mathematics] Presses Universitaires de France, Paris, 1985. 397 pp.


