I want to solve ex.1 from pag. 87 "Differential Topology" of Vietor Guillemin and Alan Pollaek. This is the first step of a sequence of exercises that should lead to the proof of the Jordan separation theorem.
Here $X$ is a compact manifold of dimension $n-1$ and $D$ is a compact manifold of dimension $n$.
The winding number of $f$ is defined as the degree modulo 2 of the map $u:X \rightarrow S^{n-1}$ defined as $u(x)=\frac{f(x)-z}{|f(x)-z|}$, that is the cardinality of the (finite) preimage of an arbitrary directional vector $\#u^{-1}(v)\mod2$. Previous results ensure this number to not depend on the choice of $v$.
What I've tried
$F(D)$ is compact, so there is a convex set, like a ball containing it. If $z$ lies outside this ball we can choice an arbitrary $x'$ such that $F$ is transversal to the line $l(t)=z + t(f(x')-z)$, from the "Boundary Theorem" the number of intersection point between this line and $f(X)\mod 2$ is equal to $0$. But this is precisely $\#u^{-1}(v) \mod 2$ for $v=\frac{f(x')-z}{|f(x')-z|}$.
What I miss
$F(D)$ is not necesseraly convex, so for an arbitrary $z$ and an arbitrary transversal line $l(t)=z + t(f(x')-z)$ it can hit $f(X)$ in both the directions, this lead to $\#u^{-1}(v) + \#u^{-1}(-v)\mod 2=0$ that does not add anything new.
The solution was immediate following the textbook.
The boundary theorem ensure that if a function on the boundary $X$ can be extended to the whole manifold $D$ its degree modulo $2$ is $0$. As $z$ is not in the image $F(D)$ the function $u(x)$ can be extended (its denominator is always different from $0$).