The degree of a map $f : S^n \to S^n$

For $n > 0$, $H_n(S^n) = \mathbb {Z}$; let $g$ denote a generator. If $f : S^n \to S^n$ is continuous, then $f_*(g) = dg$ for some integer $d$; the degree of $f$ is defined to be the integer $d = \deg (f)$ with $f_*(g) = \deg (f) g$.

Prove the following elementary facts about degree:

• $\deg (1) = 1$;

• $\deg (hf) = \deg (h)\deg (f)$;

• if $f$ is homotopic to $f'$, then $\deg (f) = \deg (f')$;

• if $f$ is a homotopy equivalence, then $\deg (f) = \pm 1$.

Also show that the degree does not depend on which generator of $H_n(S^n) = \mathbb {Z}$ you choose.

When $f : (D^{n+1}, S^n) \to (D^{n+1}, S^n)$, there is a definition of degree analogous to that of Exercise 6.9.1, using $H_{n+1}(D^{n+1}, S^n) \simeq \mathbb {Z}$: if $g$ is a generator, degree is defined by the same formula, $f_*(g) = \deg (f) g$.

Show that $\deg (f)$ as defined this way is the same as $\deg (f|_{S^n})$.

Suppose $f : (S^{n+1}, S^{n+1}_+, S^{n+1}_-, S^n) \to (S^{n+1}, S^{n+1}_+, S^{n+1}_-, S^n)$. We can define $\deg (f)$ and $\deg (f|_{S^n})$. Show that they are equal.

There is a reflection of any sphere of dimension greater than or equal to $0$ given by $r_1 : S^n \to S^n$, where $r_1(x_1, x_2, \ldots , x_{n+1}) = (-x_1, x_2, \ldots , x_{n+1})$. By starting with the case of $S^0$, show that the degree of $r_1$ is $-1$ for all $n$.

Suppose $r : S^n \to S^n$ is a reflection, of the form $r(x) = x - 2(x \cdot v)v$, where $v \in S^n$ is the normal vector of the plane through which one is reflecting (for example, for $r_1$ of Exercise 6.9.4, $v = e_1$; for $r_2(x_1, x_2, \ldots , x_{n+1}) = (x_1, -x_2, \ldots , x_{n+1})$, $v = e_2$). Let $n > 0$.

The antipodal map $A : S^n \to S^n$ is $A(x) = -x$; it can be written as the composition of $n+1$ reflections in the individual coordinates (as in Exercise 6.9.5). Show that the degree of the antipodal map of $S^n$ is $(-1)^{n+1}$.

Show that the antipodal map of $S^n$ is homotopic to the identity iff $n$ is odd.

Show that the local degree (as defined in the remark below) is well defined, independent of which disk $D_x$ is chosen.

Show that the following alternative definition of local degree (as defined in the remark to Exercise 6.9.8) gives the same answer: choose a small disk $D_y$ about $y$ so that $f(D_x) \subset D_y$ and $f : (D_x, D_x \setminus \left\{ x\right\} ) \to (D_y, D_y \setminus \left\{ y\right\} )$. Then define $\deg_x(f)$ by $f_*(g_x^D) = \deg_x(f) g_y^D$.

Show that the local degree (Exercise 6.9.8) has the following properties analogous to the degree:

• $\deg_x(1) = 1$;

• if $f(x) = y$, $h(y) = z$, and there are disks $D_x, D_y, D_z$ with $f : (D_x, D_x \setminus \left\{ x\right\} ) \to (D_y, D_y \setminus \left\{ y\right\} )$, $h : (D_y, D_y \setminus \left\{ y\right\} ) \to (D_z, D_z \setminus \left\{ z\right\} )$, then $\deg_x(hf) = \deg_y(h) \deg_x(f)$;

• if $f$ is a local homeomorphism at $x$ so that there is a disk $D_x$ which is sent homeomorphically to a disk $D_y$, then $\deg_x(f) = \pm 1$.

Suppose $f : (S^n, S_x) \to (S^n, S_y)$ (notation as in the remark to Exercise 6.9.8), and $f^{-1}(\left\{ y\right\} ) = \left\{ x\right\}$. Show that $f_*(g_x) = \deg (f) g_y$ and thus $\deg (f) = \deg_x(f)$.

It is usually not the case that $f^{-1}(\left\{ y\right\} ) = \left\{ x\right\}$, but there are geometric techniques to homotope $f$ so that $f^{-1}(\left\{ y\right\} )$ is a finite number of points $x_1, \ldots , x_k$; choosing small disks about these points, one can define the local degrees $\deg_{x_i}(f)$ as in Exercise 6.9.8. Theorem 6.9.1 states: when $f^{-1}(\left\{ y\right\} ) = \left\{ x_1, \ldots , x_k\right\}$, then $\deg (f) = \sum_{i=1}^k \deg_{x_i}(f)$.

Note that the hypothesis implies that $f : (S^n, S^n \setminus \left\{ x_1, \ldots , x_k\right\} ) \to (S^n, S_y)$, and there is a commutative diagram

Show that $H_n(S^n, S^n \setminus \left\{ x_1, \ldots , x_k\right\} ) \simeq \bigoplus_{i=1}^k H_n(D_{x_i}, D_{x_i} \setminus \left\{ x_i\right\} )$ and, using this identification, $e_x(g) = \sum_{i=1}^k g_{x_i}^D$.

Use Exercise 6.9.12 and the commutative diagram given there to prove Theorem 6.9.1 (that $\deg (f) = \sum_{i=1}^k \deg_{x_i}(f)$ when $f^{-1}(\left\{ y\right\} ) = \left\{ x_1, \ldots , x_k\right\}$).

Suppose that $T : S^n \to S^n$ is a homeomorphism and $f : S^n \to S^n$ satisfies $fT(x) = f(x)$. Suppose the local degree $\deg_x(f)$ of $f$ is defined at $x$ (as in Exercise 6.9.8). Show that $\deg_{T(x)}(f)$ is also defined and

Consider the quotient map $q : S^n \to S^n/(x \sim -x) = \mathbb {RP}^n$. Compose with the map $\mathbb {RP}^n \to \mathbb {RP}^n/\mathbb {RP}^{n-1}$ which identifies $\mathbb {RP}^{n-1}$ to a point, and identify this quotient with $S^n$ (it is homeomorphic to a disk with its boundary sphere identified to a point). This composition gives a map $S^n \to S^n$. Use Exercise 6.9.14 and Theorem 6.9.1 (Exercise 6.9.13) to show that the degree of this map is $\pm (1 + (-1)^{n+1})$.