Showing that a function is injective

Question

I am trying to show that the following function is injective in some neighborhood of $(0, 0)$: $f:\mathbb R^2 \rightarrow \mathbb R^2$ given by

$$f(x, y)=(\sin(x^3)\cosh(y), \cos(x^3)\sinh(y))$$

I tried to use the inverse function theorem but $df(0, 0)$ is not invertible.

Answer 1

Let $g(x,y)=(x^3,y)$ and $h(x,y)=(\sin x\cosh y,\cos x\sinh y)$, so $f=h\circ g$.

Then $g(0,0)=(0,0)$ and $g$ is invertible, and $h$ is locally invertible at $(0,0)$ since

$\displaystyle\frac{\partial(h_1,h_2)}{\partial(x,y)}=\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(\sinh^2y+1)+\sin^2x\sinh^2y=\sinh^2y+\cos^2x$,

so $\displaystyle\frac{\partial(h_1,h_2)}{\partial(x,y)}(0,0)=1\ne0$.

Therefore $f$ is locally invertible at $(0,0)$.