Suppose that you are given the fact that for each \(n\), there is no retraction \(r: B^{n+1} \to S^n\). (This result can be proved using more advanced techniques of algebraic topology.) Prove the following:
The identity map \(i: S^n \to S^n\) is not nulhomotopic.
The inclusion map \(j: S^n \to \mathbb{R}^{n+1} - 0\) is not nulhomotopic.
Every nonvanishing vector field on \(B^{n+1}\) points directly outward at some point of \(S^n\), and directly inward at some point of \(S^n\).
Every continuous map \(f: B^{n+1} \to B^{n+1}\) has a fixed point.
Every \(n+1\) by \(n+1\) matrix with positive real entries has a positive eigenvalue.
If \(h: S^n \to S^n\) is nulhomotopic, then \(h\) has a fixed point and \(h\) maps some point \(x\) to its antipode \(-x\).
Solution
1. Part (a): \(\operatorname{id}_{S^n}\) is not nullhomotopic.Proof:
1.1. Suppose for contradiction that \(\operatorname{id}_{S^n}\) is nullhomotopic, so there is a homotopy \(H: S^n \times I \to S^n\) with \(H(x, 0) = x\) and \(H(x, 1) = p_0\) for some \(p_0 \in S^n\).
1.2. Define \(r: B^{n+1} \to S^n\) in spherical coordinates by:\[r(t x) = H(x, 1 - t) \quad \text{for } x \in S^n, \, t \in [0, 1].\]
1.3. Because \(r(0) = H(x, 1) = p_0\) is independent of \(x\), \(r\) is continuous on all of \(B^{n+1}\).
1.4. For \(x \in S^n\) (\(t = 1\)), \(r(x) = H(x, 0) = x\).
1.5. Thus \(r\) is a retraction \(B^{n+1} \to S^n\), contradicting the no-retraction theorem.
2. Part (b): The inclusion \(j: S^n \hookrightarrow \mathbb{R}^{n+1} \setminus \{0\}\) is not nullhomotopic.Proof:
2.1. The radial projection \(\rho: \mathbb{R}^{n+1} \setminus \{0\} \to S^n\) given by \(\rho(y) = y / \|y\|\) is continuous and satisfies \(\rho \circ j = \operatorname{id}_{S^n}\).
2.2. If \(j\) were nullhomotopic, the composition \(\rho \circ j = \operatorname{id}_{S^n}\) would be nullhomotopic, contradicting Part (a).
3. Part (c): Nonvanishing vector fields point directly inward and outward.Proof:
3.1. Let \(v: B^{n+1} \to \mathbb{R}^{n+1} \setminus \{0\}\) be a continuous non-vanishing vector field.
3.2. Suppose \(v(x)\) never points directly outward on \(S^n\), meaning there is no \(x \in S^n\) and \(\lambda > 0\) such that \(v(x) = \lambda x\).
3.3. For each \(x \in B^{n+1}\), the ray starting at \(x\) in direction \(-v(x)\) intersects the boundary \(S^n\) at a unique point \(r(x) = x - t v(x)\) with \(t \ge 0\).
3.4. Since \(v(x)\) never points outward on \(S^n\), \(r(x) = x\) for \(x \in S^n\), making \(r: B^{n+1} \to S^n\) a continuous retraction, contradiction.
3.5. Replacing \(v\) with \(-v\) proves that \(v\) must also point directly inward at some point of \(S^n\).
4. Part (d): Brouwer Fixed Point Theorem.Proof:
4.1. Suppose \(f: B^{n+1} \to B^{n+1}\) has no fixed points (\(f(x) \neq x\) for all \(x\)).
4.2. For each \(x \in B^{n+1}\), the ray from \(f(x)\) through \(x\) intersects \(S^n\) at a unique point \(r(x) = x + t(x)(x - f(x))\) where \(t(x) \ge 0\).
4.3. By the quadratic formula, \(t(x)\) is continuous, and for \(x \in S^n\), \(r(x) = x\).
4.4. Thus \(r: B^{n+1} \to S^n\) is a continuous retraction, contradiction.
5. Part (e): Perron-Frobenius positive eigenvalue.Proof:
5.1. Let \(A\) be an \((n+1) \times (n+1)\) matrix with strictly positive entries \(a_{ij} > 0\).
5.2. Let \(\Delta^n = \{x \in \mathbb{R}^{n+1} \mid x_i \ge 0, \, \sum_{i=1}^{n+1} x_i = 1\}\), which is homeomorphic to \(B^n\).
5.3. For \(x \in \Delta^n\), \(x \neq 0\), so \((Ax)_i = \sum_j a_{ij} x_j > 0\) for all \(i\).
5.4. Define \(g: \Delta^n \to \Delta^n\) by \(g(x) = \frac{Ax}{\|Ax\|_1}\).
5.5. By Brouwer’s Fixed Point Theorem on \(\Delta^n \cong B^n\), \(g\) has a fixed point \(x_0 \in \Delta^n\).
5.6. Then \(Ax_0 = \lambda x_0\) where \(\lambda = \|Ax_0\|_1 > 0\), so \(\lambda\) is a positive eigenvalue.
6. Part (f): Fixed points and antipodal points of nullhomotopic maps on \(S^n\).Proof:
6.1. If \(h(x) \neq -x\) for all \(x \in S^n\), then the straight-line homotopy \(H_1(x, t) = \frac{(1-t)h(x) + t x}{\|(1-t)h(x) + tx\|}\) is well-defined, showing \(h \simeq \operatorname{id}_{S^n}\). Since \(h\) is nullhomotopic, this would force \(\operatorname{id}_{S^n}\) to be nullhomotopic, contradicting (a). Thus \(h(x) = -x\) for some \(x\).
6.2. If \(h(x) \neq x\) for all \(x \in S^n\), then \(H_2(x, t) = \frac{(1-t)h(x) - tx}{\|(1-t)h(x) - tx\|}\) shows \(h \simeq a\), where \(a(x) = -x\) is the antipodal map.
6.3. Then \(\operatorname{id}_{S^n} = a \circ a \simeq a \circ h \simeq a \circ c_{p_0} = c_{-p_0}\), making \(\operatorname{id}_{S^n}\) nullhomotopic, contradiction. Thus \(h(x) = x\) for some \(x\). Q.E.D.