We define two smooth maps $f: (\mathbb R, 0) \to (\mathbb R^2, 0)$ and $g: (\mathbb R, 0) \to (\mathbb R^2, 0)$ to be equivalent if there exist diffeomorphisms $\tau : \mathbb R \to \mathbb R$ and $\mu: \mathbb R^2 \to \mathbb R^2$ such that
$$ \mu \circ f = g \circ \tau$$
I wanted to provide an example of non equivalent maps.
Please could someone tell me if my example is correct?
Let $f(x) = (0,0)$ be the constant zero map and $g(x) = (x,x)$ the ''diagonal inclusion''.
To see that $f$ and $g$ are not equivalent assume they were. Then there are diffeomorphisms $\tau, \mu$ such that
$$ \mu \circ f = g \circ \tau$$
Then taking the derivative we get
$$ (D\mu) (Df) = (Dg)(D\tau)$$
where $D$ denotes the Jacobian. Since $\tau, \mu$ are diffeomorphisms, $D\tau$ and $D\mu$ are invertible. But because $Df=0$ the LHS is zero while the RHS is not. A contradiction.
The same, but possibly simpler: Since $f$ is constant, so is $\mu\circ f$. Since $g$ is injective, this implies that $\tau$ is constant.