Topological degree of an injective mapping is $\pm 1$

320 Views Asked by At

I'm reading the book Finite Dimensional Variational Inequalities and Complementarity Problems (Vol. 1) of Francisco Facchinei and Jong-Shi Pang. They defined the topological degree axiomatically as follow ($\Gamma$ is the collection of triples $(\Phi, \Omega, p)$ where $\Omega$ is a bounded open subset of $\mathbb{R}^n$, $\Phi$ is a continuous mapping from $\overline{\Omega}$ to $\mathbb{R}^n$ and $p\not \in \Phi(\partial \Omega)$):

enter image description here

Then they presented a proposition which "can be viewed as a generalization of Axiom (A1)":

Let $\Omega$ be a nonempty, bounded open subset of $\mathbb{R}^n$ and let $\Phi\colon \overline{\Omega}\to \mathbb{R}^n$ be a continuous injective mapping. For every $p\in \Phi(\Omega)$, $\text{deg}(\Phi, \Omega, p) = \pm 1$.

I do not know how to prove this proposition. I tried to construct a homotopy between $\Phi$ and $\text{id}$ (but I could not), and the part $\pm 1$ made me confused.

Thank you very much for any hint or solution.