This question is about the book Topology from the Differentiable Viewpoint of Milnor.
Let $M$ and $N$ be oriented $n$-manifolds without boundary, and assume $M$ is compact and $N$ is connected. Let $x\in M$ be a regular point of $f$, so that $df_X:TM_x\to TN_{f(x)}$ is a vector space isomorphism. Define the sign of $df_x$ to be $+1$ or $-1$ according as $df_x$ preserves or reverses orientation.
How can we show that the sign of $df_x$ is locally constant function of $x$?
Since $M$ is oriented, $x$ has a neighborhood $U$ and a diffeomorphism $h$ of $U$ onto an open subset $V$ of $\Bbb R^n$ which is orientation preserving, in the sense that for each $y\in U$ the isomorphism $dh_y$ carries the specified orientation of $TM_y$ to the standard orientation of $\Bbb R^n=TV_{h(y)}$.
I think I should use this fact, but I can't see how does this imply that the sign of $df_x$ is constant in a neighborhood of $x$.
Best Answer
At some points Milnor's phantastic book is a "little short". I think you found one of these.
If $x$ is a regular value of $f$, then $f$ maps an open neighborhood $U$ of $x$ in $M$ diffeomorphically onto an open subset $V \subset N$. By shrinking $U$ and $V$, we may assume that there are orientation preserving diffeomorphisms $g : U \to U'$ and $h : V \to V'$, where $U', V' \subset \mathbb R^n$ are open. By shrinking $U'$, we may assume that $U'$ is an open ball in $\mathbb R^n$. Now consider
$$\phi = h \circ f\mid_{U} \circ g^{-1} : U' \to V' .$$ This is a diffeomorphism. Let $D\phi_y : \mathbb R^n \to \mathbb R^n$ denote the ordinary derivative of $\phi$ at $y \in U'$ in the sense of multivariable calculus (which is the best linear approximation of $\phi$ at $y$). Then we have $\text{sign} df_{g(y)} = \text{sign} D\phi_y$. It therefore suffices to show that $\text{sign} D\phi_y$ is constant on $U'$.
Although Milnor doesn't mention it, it is clear that $\text{sign} D\phi_y$ is nothing else then the sign of the determinant $\det D\phi_y$. Thus it suffices to show that $\det D\phi_y$ does not change its sign on $U'$.
We know that $\phi$ is smooth, in particular continuously differentiable. Thus the map $$D\phi : U' \to Hom(\mathbb R^n,\mathbb R^n), y \mapsto D\phi_y$$ is continuous. Here $Hom(V,W)$ denotes the vector space of all linear maps $V \to W$ with topology induced by any norm. It is well-known that the determinant is a continuous function $\det : Hom(\mathbb R^n,\mathbb R^n) \to \mathbb R$. The function $\psi =\det \circ D\phi : U ' \to \mathbb R$ does not have a zero because all $D\phi_y$ are invertible. Since $U'$ is connected, $\psi(U')$ is a connected subset of $\mathbb R$ - i.e. an interval - not containing $0$. This means that the sign of $\psi(y)$ does not change on $U'$.