Let $M^{n-1} \subset \mathbb{R}^n$ be a smooth compact manifold, $\Lambda$ = $\{U_i\}_{i \in I}$ be an open cover of $\mathbb{R}^n$. Suppose that for every $U_i$ there is a smooth bounded function $f_i : U_i \rightarrow [-1,1]$, such that, for every $i \neq j$, occurs only one of the following conditions:
- $f_i(x) = f_j(x)$, $\forall$$x$$\in$$U_i \cap U_j,$
- $f_i(x) = -f_j(x)$, $\forall$$x$$\in$$U_i \cap U_j.$
Besides that $f_i(x) = 0$$\Leftrightarrow$$x \not\in M\cap U_i$.
For $x_0$$\in$$\mathbb{R}^n \setminus \{0\}$, we will define a function $f_{x_0}:\mathbb{R}^n \rightarrow [-1,1]$ as following:
Consider $x$$\in$$\mathbb{R}^n$, let $\gamma_x:[0,1] \to \mathbb{R}^n$ be a continious path linking $x_0$ to $x$. Note that we can subdivide $[0,1]$ in $0=t_0\leq t_1 \leq t_2 \leq \ldots \leq t_n =1$, in such way that $\gamma_x([t_i,t_{i+1}]) \subset U_{j_i}$, for some $U_{j_i}$$\in$$\Lambda$. Suppose, without loss of generality, that $U_{j_i} = U_i$, for every $i$$\in$$\{0,1,...,n\}$.
For every $i$$\in$$\{0,...,n\}$ there is $\xi_i$$\in$$U_{i-1} \cap U_{i}$, such that, $f_{i}(\xi_i) \neq 0.$
Defining $g_i:U_i \rightarrow [-1,1]$ recursively as
- $g_{0}(x)$$=$$f_{0}(x)$,
- $g_i (x)= \frac{f_i(\xi_i)}{g_{i-1}(\xi_{i})} \cdot f_i(x)$, $\forall$$i$$\in\{1,...,m\}$.
Finally we say that $f_{x_0}(x) = g_n(x)$, for $x \in U_n$.
My Doubt: Why $f_{x_0}$ does not depends on the path $\gamma_x$ used to define $f_{x_0} (x)$?
I think the function $f$ is well defined because $\mathbb{R}^{n}$ is a simply connected, but I don't know how to demonstrate that the construction of the value $f_{x_0}(x)$ is invariant under homotopy.