Let $\Omega =B_1(0)=\{z\in\mathbb{C}\cong\mathbb{R}^2 : \lvert z\rvert<1\}, y=0,$ and define
$h(t,z)=\begin{cases} \lvert z\rvert, & t=0, z\in\overline\Omega \\ \lvert z\rvert \exp(i\phi/t), & 0<t\leq1, z=\lvert z\rvert e^{i\phi}, 0\leq\phi\leq2\pi t \\ \lvert z\rvert, & 0<t<1, z=\lvert z\rvert e^{i\phi}, 2\pi t<\phi\leq2\pi \end{cases}$
We have that $h(t,\cdot)$ and $h(\cdot,z)$ are continuous and $h(t,z)\neq0$ on $[0,1]\times\partial\Omega$ and $d\big(h(t,\cdot),\Omega,0\big)=1$, where $d$ is the topological degree. We also have that $h(0,\dot)$ is homotopic to $f(z)\equiv(1,0)$ by $g(s,z)=s(\lvert z\rvert,0)+(1-s)(1,0)$. Why is it not true that $d(h(0,\cdot),\Omega,0)=0$ (if that's true we have contradiction with the homotopy invariance of the topological degree, a contradiction)? Do I have to show that the homotopy is not legal, that's it $y\in g(0,\partial\Omega)?$