I was reading Milnor's differential topology book in page 27,there is a statement :
Let $f:X\to Y$ be smooth map with $X$ and $Y$ be same dimension,such that both of them are oriented , we further assume $X$ is compact and $Y$ is connected.
denote the Brouwe degree by $$\text{deg}(f)_y = \sum_{x \in f^{-1}(y)} \text{sgn}(\det(df_x))$$ prove it does not depend on the regular value $y$ we choose.in other words deg as function of y is constant.
The idea is :to prove that deg is locally constant,then by connectess it's contant everywhere
since we know for the setting above,any regular value $y$.in fact exist a pile of neiborhood $V_1,V_2,...V_n \subset M$ such that diffeomorphic to neiborhood of $y$， that is exist local diffeomorphism $V_i\to U_y$,as y varies in $U_y$ will not change the number of term in the sum
the question is why $ \text{sgn}(\det(df_x))$ will also not change as y varies in $U_y$,the intuition is easy that is as $y$ change a little bit,$x$ will change a liile bit,hence det will not change too much to filp the sign How to prove it ?
It's far simpler. If $f$ maps each $V_i$ diffeomorphically to $U$, then $df_x$ is an isomorphism from $T_xV_i$ to $T_{f(x)}U$ for every $x\in V_i$. In particular, $df_x$ is either orientation-preserving for all $x\in V_i$ or orientation-reversing for all $x\in V_i$. (If you choose a parametrizations/charts, then you can look at the square matrices, and their determinants vary continuously; one cannot change from positive determinant to negative determinant without passing through $0$, but this would violate the fact that the matrices are all non-singular.)