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

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 $p\not \in \Phi(\partial \Omega)$): 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.

• Hint: Use a non-axiomatic definition of the degree. (You can find one, for instance, in the book by Guillemin and Pollack, "Differential Topology". Or, just check Wikipedia: en.wikipedia.org/wiki/…) Then this proposition becomes a tautology. – Moishe Kohan Dec 15 '17 at 12:06
• Maybe they wanted you to prove this proposition just using axioms A1-A_3. I doubt this is possible (at least, not without some very serious work). The point is that while any embedding of $\Omega$ into $R^n$ is homotopic to the identity, they are not isotopic. In particular, one can find easy examples when any homotopy will have to send some boundary points of $\Omega$ to $p$. Of course, one can subdivide $\Omega$ into a union of finitely many balls (assuming that $\Omega$ is nice enough). It is true that any embedding $B^n\to R^n$ is either isotopic to the identity map or – Moishe Kohan Dec 15 '17 at 16:03
• to a hyperplane reflection, but this is a nontrivial topological result. – Moishe Kohan Dec 15 '17 at 16:03