# Does the homeomorphism have a non-negative or non-positive determinant?

Let $$\Omega_1$$ and $$\Omega_2$$ be domains (open and connected) in $$\mathbb{R}^2$$. $$\psi:\Omega_1\to\mathbb{R}$$ and $$\phi:\Omega_1\to\mathbb{R}$$ are $$C^1$$ functions with two variables. Moreover, we assume that map $$(x,y)\to (\phi(x,y),\psi(x,y))$$ is homeomorphism from $$\Omega_1$$ to $$\Omega_2$$, i.e. the map $$(x,y)\to (\phi(x,y),\psi(x,y))$$ is continuous from $$\Omega_1$$ to $$\Omega_2$$ and has continuous inverse map. I want to ask that if I can obtain that the Jacobi determinant of the map, denoted as $$\frac{\partial(\phi,\psi)}{\partial(x,y)}$$ is either non-positive or non-negative in $$\Omega_1$$, i.e. either $$\frac{\partial(\phi,\psi)}{\partial(x,y)}\geq 0$$ for all $$(x,y)\in\Omega_1$$, or $$\frac{\partial(\phi,\psi)}{\partial(x,y)}\leq 0$$ for all $$(x,y)\in\Omega_1$$. I have tried by considering the image sets of a curve in $$\Omega_1$$ and by using the connectedness of $$\Omega_1$$ but failed. Can you give me some references or hints?

Let $$f=(\phi,\psi):\Omega_1\to\Omega_2$$.
For every point $$p\in\Omega_1$$ consider the curve $$\gamma:t\mapsto p+\varepsilon e^{it}$$, for $$\varepsilon$$ so small that the curve is contained in $$\Omega_1$$. Let $$n(p)$$ be the winding number of $$f\circ \gamma$$ around $$f(p)$$. As $$f$$ is a homeomorphism, $$n(p)$$ is well defined (and an integer) and depends continuously on $$p$$, so as $$\Omega_1$$ is connected, it is constant. So your statement is a direct consequence of the fact that $$n(p)=1$$ in points $$p$$ where $$\frac{\partial(\phi,\psi)}{\partial(x,y)}(p)>0$$ and $$n(q)=-1$$ in points $$q$$ where $$\frac{\partial(\phi,\psi)}{\partial(x,y)}(q)<0$$.
This can be generalized to higher dimensions using the local degree of $$f$$ instead of winding numbers.