Let \(p(z) = \sum_{j=0}^n a_j z^j\) be a polynomial of degree \(n \ge 1\) with \(a_n \neq 0\). Extend \(p\) to a continuous self-map of the Riemann sphere \(\widehat{\mathbb{C}} = \mathbb{C} \cup \{\infty\} \cong S^2\) by setting \(p(\infty) = \infty\). Determine the topological degree \(\deg(p)\).
Solution
Goal: Let \(p(z) = \sum_{j=0}^n a_j z^j\) with \(a_n \neq 0\) and \(n \ge 1\) be a polynomial with complex coefficients. Extend \(p\) to a continuous map on the Riemann sphere \(\widehat{\mathbb{C}} = \mathbb{C} \cup \{\infty\} \cong S^2\) by setting \(p(\infty) = \infty\). Determine the topological degree \(\deg(p)\). (If \(n = 0\), \(p(z) \equiv a_0\) is constant).
1. Case \(n = 0\) (constant polynomial \(p(z) = a_0\)). Then \(\deg(p) = 0\).
1.1. If \(p\) is constant, its image is the single point \(\{a_0\}\).
1.2. A non-surjective map \(g \colon S^2 \to S^2\) factors through \(S^2 \setminus \{q\} \simeq \mathbb{R}^2\) for some \(q \notin g(S^2)\).
1.3. Since \(\mathbb{R}^2\) is contractible, \(H_2(\mathbb{R}^2) = 0\), so the induced map \(g_* \colon H_2(S^2) \to H_2(S^2)\) factors through the zero group and is the zero homomorphism.
1.4. Hence \(\deg(g) = 0\).
2. Case \(n \ge 1\) with \(a_n \neq 0\). Then \(\deg(p) = n\).
2.1. The extension to \(\widehat{\mathbb{C}}\) is continuous: Since \(a_n \neq 0\) and \(n \ge 1\), \(\lim_{|z| \to \infty} |p(z)| = \lim_{|z| \to \infty} |a_n z^n| \left|1 + \sum_{j=0}^{n-1} \frac{a_j}{a_n z^{n-j}}\right| = \infty\), so setting \(p(\infty) = \infty\) makes \(p \colon \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\) continuous.
2.2. Construct a homotopy to \(q(z) = a_n z^n\): Define \(H \colon \widehat{\mathbb{C}} \times [0, 1] \to \widehat{\mathbb{C}}\) by \[H(z, t) = a_n z^n + (1-t)\sum_{j=0}^{n-1} a_j z^j \quad \text{for } z \in \mathbb{C}, \qquad H(\infty, t) = \infty.\]
2.3. For each \(t \in [0, 1]\), \(H(z, t)\) is a polynomial of degree \(n\) with leading coefficient \(a_n \neq 0\), so \(\lim_{|z| \to \infty} |H(z, t)| = \infty\) uniformly for \(t\). Thus \(H\) is a continuous homotopy of maps \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\).
2.4. The degree of a map \(S^2 \to S^2\) is determined by its induced map on \(H_2(S^2) \cong \mathbb{Z}\). Since homotopic maps induce the same homomorphism on homology, \(\deg(p) = \deg(z \mapsto a_n z^n)\).
3. Homotopy to the standard monomial \(z \mapsto z^n\).
3.1. Write \(a_n = r e^{i\theta}\) with \(r > 0\) and \(\theta \in \mathbb{R}\).
3.2. Define \(K \colon \widehat{\mathbb{C}} \times [0, 1] \to \widehat{\mathbb{C}}\) by \(K(z, t) = ((1-t)r + t) e^{i(1-t)\theta} z^n\) for \(z \in \mathbb{C}\) and \(K(\infty, t) = \infty\).
3.3. Since \((1-t)r + t > 0\) for all \(t \in [0, 1]\), the leading coefficient is non-zero for all \(t\), so each \(K(\cdot, t)\) is a polynomial of degree \(n\) extending continuously to \(\widehat{\mathbb{C}}\).
3.4. Moreover \(K(z, 0) = r e^{i\theta} z^n = a_n z^n\) and \(K(z, 1) = z^n\), so \(K\) is a homotopy from \(z \mapsto a_n z^n\) to \(f_n(z) = z^n\).
3.5. Homotopic maps have the same degree, so \(\deg(z \mapsto a_n z^n) = \deg(f_n)\).
4. Compute the degree of \(f_n(z) = z^n\) on \(\widehat{\mathbb{C}} \cong S^2\).
4.1. Method 1 (Local degree / Regular values): Choose the regular value \(w_0 = 1 \in \mathbb{C} \setminus \{0\}\). The preimage \(f_n^{-1}(1) = \{\zeta_k = e^{2\pi i k / n} \mid k = 0, 1, \dots, n-1\}\) consists of \(n\) distinct roots of unity.
4.2. At each preimage \(\zeta_k\), \(f_n'(\zeta_k) = n \zeta_k^{n-1} \neq 0\). Since \(f_n\) is complex holomorphic, its Jacobian determinant as a real map \(\mathbb{R}^2 \to \mathbb{R}^2\) is \(|f_n'(z)|^2 > 0\), so each local degree is \(\deg_{\zeta_k}(f_n) = +1\).
4.3. By the regular value formula, the degree equals the sum of local degrees over the preimage of any regular value: \(\deg(f_n) = \sum_{k=0}^{n-1} (+1) = n\).
4.4. Equivalently, one may first homotope \(f_n\) through maps fixing \(0\) and \(\infty\) to a latitude-preserving model whose restriction to each nonpolar latitude circle is \(e^{i\theta}\mapsto e^{in\theta}\). That latitude-preserving model is the suspension of the degree-\(n\) circle map, so naturality of the suspension isomorphism again gives degree \(n\). The local-degree computation in <2>1–<2>3 already proves the result without this additional homotopy.
5. Conclusion.
5.1. For \(p(z) = \sum_{j=0}^n a_j z^j\) with \(a_n \neq 0\): \[\deg(p) = n.\] (If \(p\) is constant, \(\deg(p) = 0\)).
5.2. By 2 and 3, \(\deg(p) = \deg(f_n)\); by 4, \(\deg(f_n) = n\); and by 1, the constant case has degree \(0\).