For \(n\geq 1\) and a continuous map \(f\colon S^n\to S^n\), the degree \(\deg f\in\ZZ\) is defined by \[ f_*[S^n]=\deg(f)\,[S^n]\in H_n(S^n;\ZZ). \]
The degree
(Properties of the degree)
Let \(n\geq 1\) and \(f,g\colon S^n\to S^n\).
- A constant map has degree \(0\).
- \(\deg\id_{S^n} = 1\).
- \(\deg (f\circ g) = \deg f \cdot \deg g\).
- If \(f\homotopic g\), then \(f_* = g_*\), so \(\deg f = \deg g\). Conversely, by Hopf’s theorem, maps of equal degree are homotopic.
- If \(f\) is a homotopy equivalence with homotopy inverse \(h\), then \(\deg f\cdot\deg h = \deg\id = 1\), so \(\deg f = \pm 1\).
- For \(1\leq i\leq n+1\), the reflection \(R_i\colon S^n\to S^n\), \((x_1, \ldots, x_i, \ldots, x_{n+1}) \mapsto (x_1, \ldots, - x_i, \ldots, x_{n+1})\), has degree \(-1\).
- The antipodal map \(x\mapsto -x\) on \(S^n\subseteq \RR^{n+1}\) is the composite \(R_1\circ\cdots\circ R_{n+1}\), so it has degree \((-1)^{n+1}\). In particular, a map of even degree is not homotopic to the antipodal map.
Exercises
No fixed points implies homotopic to antipodal (Tag E-EYILL)
Show that if \(f: S^n\to S^n\) has no fixed points \(\iff \deg f = (-1)^{n+1}\) and \(f\) is homotopic to the antipodal map.
If \(f(x)\neq x\), the line segment \(L(-x, f(x))\) does not contain \(0\).
If \(f(x) \neq -x\), the line segment \(L(x, f(x))\) does not contain \(0\).
Solution
The straight line homotopy \(H(t, x) = (1-t)f(x) + t(-x)\) is a homotopy between \(f\) and the antipodal map in \(\RR^{n+1}\). Use that \(H(t, x) = 0 \iff t=1/2 \iff f(x) = x\), so \(H(t, \wait)\) is always a line from \(x\) to \(f(x)\) not passing through \(0\), so \(H(t, \wait) \neq 0\). So this descends to a homotopy of \(\RR^{n+1}\smz \homotopic S^n\), or explicitly one can project \[ H: I\cross S^n &\to S^n \\ (t, x) &\mapsto {H(t, x) \over \norm{H(t, x)}} .\]
Less explicitly: if \(f(x) \neq x\) for all \(x\in S^n\), there is a unique geodesic through these two points, so let each point flow along its corresponding geodesic.
Degree of a non-surjective map $S^n\to S^n$ is zero (Tag E-ZXKDY)
Show that if \(f\) is not surjective then \(\deg f = 0\).
Solution
Choose \(p\in S^n\setminus f(S^n)\). Then \(f\) factors as \[ S^n \mapsvia{f_1} S^n\smts{p}\mapsvia{j} S^n, \] where \(j\) is the inclusion. Since \(S^n\smts{p}\cong\RR^n\) is contractible, \[ H_n(S^n\smts{p})=0. \] Hence the induced map on top homology factors through the zero group: \[ H_n(S^n)\mapsvia{(f_1)_*}0\mapsvia{j_*}H_n(S^n). \] Thus \(f_*=0\) on \(H_n(S^n)\cong\ZZ\). By the definition of degree, \(\deg f=0\).
Brouwer and Lefschetz
Brouwer fixed-point theorem (Tag T-S2OLJ)
Every continuous map \(f\colon D^n \to D^n\) has a fixed point [@Hat02].
(Brouwer's theorem via retractions)
Let \(n\geq 1\) and suppose \(f\colon D^n \to D^n\) has no fixed point. For \(x\in D^n\), let \(g(x)\) be the point where the ray from \(f(x)\) through \(x\) meets \(\del D^n = S^{n-1}\). Then \(g\colon D^n\to S^{n-1}\) is continuous, and \(g(x)=x\) for \(x\in S^{n-1}\), so with \(\iota\colon S^{n-1}\injects D^n\) the inclusion, \(g\circ\iota = \id_{S^{n-1}}\). On reduced homology in degree \(n-1\) this gives \(\id = g_*\circ\iota_*\), which factors through \(\tilde H_{n-1}(D^n) = 0\). This contradicts \(\tilde H_{n-1}(S^{n-1})\cong\ZZ\).
Lefschetz fixed point theorem (Tag T-BX4LD)
Let \(X\) be a finite simplicial complex, or a retract of one, and let \(f\colon X\to X\) be continuous with Lefschetz number \[ \tau(f) = \sum_{k \geq 0} (-1)^k \tr\qty{f_*\colon H_k(X; \QQ) \to H_k(X; \QQ)} .\] If \(\tau(f) \neq 0\), then \(f\) has a fixed point [@Hat02].
Brouwer’s theorem is also the case \(X = D^n\) of the Lefschetz fixed point theorem: \(D^n\) is contractible, so \(H_0(D^n;\QQ) = \QQ\), all higher homology vanishes, and every self-map \(f\) has Lefschetz number \(\Lambda_f = \operatorname{Tr}(f_* \mid H_0) = 1 \neq 0\). The implication \(\Lambda_f\neq 0\Rightarrow\) \(f\) has a fixed point has no converse in general: \(\Lambda_f=0\) does not imply that \(f\) is homotopic to a map without fixed points.
Borsuk–Ulam
Borsuk--Ulam theorem (Tag T-WNOWY)
For every continuous map \(f\colon S^n \to \RR^n\) there is a point \(x\in S^n\) with \(f(x) = f(-x)\) [@Hat02].
(The case $n=1$)
For continuous \(f\colon S^1\to\RR\), set \(g(x)\coloneqq f(x)-f(-x)\). Then \(g(-x)=-g(x)\), so \(g\) takes values of both signs or vanishes, and the intermediate value theorem on a path from \(x\) to \(-x\) gives \(x\) with \(g(x)=0\), that is, \(f(x)=f(-x)\).
(Ham sandwich theorem)
For Lebesgue measurable sets \(A_1,\ldots,A_n\subseteq\RR^n\) of finite measure, there is an affine hyperplane \(P\) such that each \(A_i\) meets the two open half-spaces of \(\RR^n\sm P\) in sets of equal measure.
For \(u=(u_0,u')\in S^n\subseteq\RR\times\RR^n\), let \(f_i(u)\) be the measure of \(A_i\cap\ts{x \st \inner{u'}{x} > u_0}\), which is continuous in \(u\). By the Borsuk–Ulam theorem applied to \(f=(f_1,\ldots,f_n)\colon S^n\to\RR^n\), there is \(u\) with \(f(u)=f(-u)\). If \(u'=0\), then one of the two sets is empty and the other is \(\RR^n\), so \(f(u)=f(-u)\) forces every \(A_i\) to have measure \(0\) and any hyperplane works. Otherwise \(P\coloneqq\ts{\inner{u'}{x}=u_0}\) is a hyperplane of measure zero, and \(f(u)=f(-u)\) says that \(P\) bisects each \(A_i\).
Hairy ball theorem
Hairy ball theorem (Tag T-VQTR6)
For \(n\geq 1\), the sphere \(S^n\) has a continuous field of nonzero tangent vectors if and only if \(n\) is odd. In particular, every continuous tangent vector field on \(S^{2k}\) vanishes somewhere [@Hat02].
Suppose \(S^k\subseteq\RR^{k+1}\) admits a nonvanishing continuous tangent vector field \(v\), so \(x \cdot v(x) = 0\) for all \(x\in S^k\). Define \(H\colon S^k \times [0, 1] \to \RR^{k+1}\) by \[ H(x, t) \coloneqq x \cos(\pi t) + \frac{v(x)}{\norm{v(x)}} \sin(\pi t). \] Since \(x \perp v(x)\), \(\norm{H(x, t)}^2 = \cos^2(\pi t) + \sin^2(\pi t) = 1\), so \(H\) is a homotopy in \(S^k\) from \(H(\wait,0)=\id_{S^k}\) to \(H(\wait,1)=-\id_{S^k}\). By homotopy invariance, \(1=\deg(\id_{S^k}) = \deg(-\id_{S^k}) = (-1)^{k+1}\), so \(k\) is odd.
Every odd-dimensional sphere \(S^{2n+1}\subseteq\CC^{n+1}\) admits the nonvanishing tangent vector field \(v(z)=iz\). The spheres \(S^0\), \(S^1\), \(S^3\), and \(S^7\) are parallelizable.