Q:

A question about the condition of Frobenius theorem

I puzzled about the condition of Frobenius theorem:
Condition FR1:
Let $X$ be a manifold, $E$ is a subbundle of $TX$,vector fields $ ξ,η $ lie in $E$(i.e. $ ξ(x),η(x)\in E_x $),then bracket $[ξ,η]$ also lie in $E$.
I am puzzled whether the condition is satisfied by any subbundle $E$ of $TX$ because of the reason stated blow .
If $E$ is a subbundle of $TX$, $U$ is sufficintly small neighborhood of $x_0 \in X$,then there exist trivizlizations
$$ 
\tau_1 : TX_U\rightarrow U\times F_1\times F_2 $$
$$
\tau_2 : E_U\rightarrow U\times F_1$$
making the following diagram commutative:
$$
$$\begin{array}
AE_U & \stackrel{}{\longrightarrow} & TX_U \\
\downarrow{\tau_2} & & \downarrow{\tau_1} \\
U\times F_1 & \stackrel{}{\longrightarrow} & U\times F_1\times F_2  
\end{array}
The bottom map is the natural one:Identity on $U$ and the injecction of $F_1$ on $F_1 \times 0  $.
If vector fields $ ξ,η $ lie in $E$,then local expression shows:
$$ξ :U\rightarrow F_1 \times 0  $$
$$η :U\rightarrow F_1 \times 0  $$
Hence 
$$[ξ,η]=η'ξ-ξ'η:U\rightarrow F_1 \times 0  $$
i.e. $[ξ,η]$ lie in $E$.
I am puzzled about it and I hope you can give an example or check out the mistakes I had made.
I will appreciate your help.

A:

I know why it happens.
Bracket $[ξ,η]$ is not invariant under VB-isomorphism.
Example by user8268 :
$X=R^3$ , $ ξ(x,y,z)=(x,y,z,0,1,0),η(x,y,z)=(x,y,z,1,0,y)$,then$[ξ,η]=(0,0,1)$.
Under VB-isomorphsim:
$$f:TR^3 \rightarrow R^3 \times R^3$$
$$f(x,y,z,m,n,l)=(x,y,z,m,n,l-ym)$$
Then $f[ξ]=(x,y,z,0,1,0),f[η)=(x,y,z,1,0,0),[fξ,fη]=0 \neq f[ξ,η]=(x,y,z,0,0,1)$.

