I am doing self-reading on some degree theory using "Topological Degree Theory and Applications" by Donal O’Regan, Yeol Je Cho, and Yu-Qing Chen; I am totally new to this field and not quite familiar with the notion. We first define the degree of a $C^1$ map as follows:
where the $J_f := \det Df$. I am stuck on this basic theorem:
I do not see how a direct computation works in this case. Could anyone help me with this? Thanks!
Assuming that the answer to my question in the comments is that $\Bbb R^m \equiv \{x \in \Bbb R^n\mid x_i = 0, \forall i > m\}$, note that $g$ carries $\overline\Omega \cap \Bbb R^m$ to $\Bbb R^m$. $g_m$ is not just the result of restricting the domain of $G$ to $\overline\Omega \cap \Bbb R^m$, but also the codomain to $\Bbb R^m$.
Now since $f$ maps into $\Bbb R_m$, the coordinate maps $f_i \equiv 0$ for all $i > m$ and therefore $\dfrac{\partial f_i}{\partial x_j} = 0$ for all $j$. So for $x\in\overline\Omega \cap \Bbb R^m$, $$Dg = \begin{bmatrix}D{g_m} & M\\\mathbf 0 & I_{n-m}\end{bmatrix}$$ for some matrix $M$. But this is enough to show $\det Dg = \det Dg_m$.