From Chapter 2 Section 5 of Guillemin and Pollack, Differential Topology,

$\mathbf{X}$ is a compact connected manifold, and $f:\mathbf{X}\rightarrow \mathbb{R}^n$ a smooth map and $\dim{X}=n-1$. (So, $f$ might be the inclusion map of a hypersurface into $\mathbb{R}^n$)

For any point $z\in\mathbb{R}^n$ define $u:\mathbf{X}\rightarrow \mathbf{S}^{n-1}$

\begin{equation} u(x) = \frac{f(x) - z}{|f(x) - z|} \end{equation}

Define the mod 2 winding number of $f$ around $z$ to be $W_2(f,z) = deg_2(u)$.

My problem is in seeing that $u$ is transversal to $\mathbf{S}^{n-1}\subset \mathbb{R}^n$.

Definition: A map $f:\mathbf{X}\rightarrow \mathbf{Y}$ is said to be transversal to a submanifold $\mathbf{Z}\subset \mathbf{Y}$ at a point $z\in f(\mathbf{X})\cap \mathbf{Z}$ if for all points $x\in f^{-1}(z)$:

\begin{equation} Image(df_x) + T_z(Z) = T_z(Y) \end{equation}

Example, let's just take $\mathbf{X} = \mathbf{S}^{1}$, the unit circle in $\mathbb{R}^2$, and $\mathbf{Y} = \mathbb{R}^2$. $T_z(Y) = \mathbb{R}^2$, but $T_z(Z) = Image(du_x)$, so how can $u$ be transversal to $\mathbf{S}^1$?


  • $\begingroup$ "Transversal" is a noun. "Transverse" is an adjective. $\endgroup$
    – anomaly
    Commented Nov 9, 2014 at 22:04

1 Answer 1


Actually, $u$ is not transversal to $\mathbf{S^{n-1}}$. The mod 2 degree of $u$ is the number of times $u$ intersects $\{z\}$, if $u$ is transvesral to $\{z\}$. If $u$ is not transversal to $\{z\}$, then we can just alter $u$ homotopically (say to some map $u_1$) to create a map that is transversal to $\{z\}$. Since homotopic maps have the same degree mod 2, the number of points in $u^{-1}_1(z)$ mod 2 is the mod 2 degree of $u$.


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