Proof. Let \(f \in C(X)\) be an \(\varepsilon\)-approximate solution of \(P\). Take any \(x_0 \in X\), and let \(w_0 := P(x_0,f(x_0))\). Since \(f\) is an \(\varepsilon\)-approximate solution of \(P\), we know that \(|w_0| < \varepsilon\). Observe that \[\begin{align} z^n + a_{n-1}(x_0)z^{n-1} + \cdots + a_1(x_0)z + (a_0(x_0) - w_0) \end{align}\] is a polynomial over \(\mathbb{C}\) possessing \(f(x_0)\) as a root. Hence by Cauchy’s bound for the root of a polynomial, we have that \[\begin{align} |f(x_0)| &\leq 1 + \max\{ |a_0(x_0) - w_0|, |a_1(x_0)| , \ldots, |a_{n-1}(x_0)| \} \\ &\leq 1 + \max\{ |a_0(x_0)| + |w_0|, |a_1(x_0)| , \ldots, |a_{n-1}(x_0)| \} \\ &\leq 1 + |w_0| + \max\{ |a_0(x_0)|, |a_1(x_0)| , \ldots, |a_{n-1}(x_0)| \} \\ &\leq 1 + \varepsilon+ \max\{ |a_0(x_0)|, |a_1(x_0)| , \ldots, |a_{n-1}(x_0)| \} \\ &\leq 1 + \varepsilon+ M \end{align}\] As the above bound holds for any \(x_0 \in X\), we obtain \(\left\Vert f \right\Vert_{\infty} \leq 1 + \varepsilon+ M\). ◻

Proof. For each \(0 \leq i \leq r\) consider the circle \(C_i = \{z \in \mathbb{C}: |z - \lambda_i| = \delta\}\), which by construction contains no root of \(p(z)\). Let \[\mu = \min_{0 \leq i \leq r} \min_{z \in C_i} |p(z)|\] where the minima are achieved and positive by compactness of the circles and continuity of \(p\). Then since the values of a polynomial over \(\mathbb{C}\) depend continuously on its coefficients, we can find an \(\varepsilon> 0\) such that if \(q(z) = z^n + b_{n - 1}z^{n - 1} + \cdots + b_1z + b_0\) is a polynomial with \(|a_k - b_k| < \varepsilon\) then \[\max_{0 \leq i \leq r} \max_{z \in C_i} |q(z) - p(z)| < \mu\] But then for any such \(q(z)\), we have \(|q(z) - p(z)| < |p(z)|\) for all \(z \in C_i\), which by Rouché’s Theorem implies that \(p(z)\) and \(q(z)\) have the same number of zeroes in the disk \(\{z \in \mathbb{C} : |z - \lambda_i| < \delta\}\) as required. ◻

Proof. Let \(R := 2 + \max_{0 \leq i < n} \left\Vert a_i \right\Vert\) and consider the set \[Y := \{(x, w) \in X \times \mathbb{C}\;: \;|w| \leq R, \quad d(w, \sigma_{P(x,z)}) \geq \delta\}\] Using Corollary 8 we see that the function \((x, w) \mapsto d(w, \sigma_{P(x,z)})\) is continuous, and hence \(Y\) is a closed set. Being a subset of the compact \(X \times \{w \in \mathbb{C} : |w| \leq R\}\), we see that \(Y\) is also compact. Let \[\varepsilon:= \min (\{ |P(x,w)| : (x,w) \in Y \} \cup \{1\})\] which is positive as \(Y\) does not contain a pair \((x, w)\) such that \(P(x,w) = 0\).

Now assume that \(f \in C(X)\) is an \(\varepsilon\)-approximate solution of \(P\). Then by Lemma 6 we get that \(|f(x)| \leq 1 + \varepsilon+ \max_{0 \leq i < n} \left\Vert a_i \right\Vert \leq R\) for all \(x \in X\). However for each \(x_0 \in X\), we have that \(|P(x_0,f(x_0))| < \varepsilon\), which by our choice of \(\varepsilon\), implies that \(|P(x_0,f(x_0))| < |P(x,w)|\) for all \((x,w) \in Y\). But this means that for each \(x_0 \in X\), the pair \((x_0, f(x_0))\) cannot be in \(Y\), and since \(|f(x_0)| \leq R\) this is only possible when \(d(f(x_0), \sigma_{P(x_0,z)}) < \delta\). ◻

Proof. Clearly if \(P\) has an exact root then it has approximate roots. Conversely assume that \(P\) has approximate roots. Let \(\delta : X \to [0,\infty)\) be the function which assigns to each \(x \in X\), the minimum distance between any two roots of of the complex polynomial \(P(x,z)\), i.e. \[\begin{align} \delta(x) = \min\{ |\lambda - \mu| : \lambda, \mu \in \sigma_{P(x,z)}, \lambda \neq \mu \} \end{align}\] Observe that \(\delta\) is non-vanishing since \(P\) has no repeated roots, and it is a continuous function by Lemma 7. In particular, it achieves a minimum since \(X\) is compact; let \(\alpha := \frac{1}{2}\min_{x \in X} \delta(x)\). By Lemma 9, we can find an \(\varepsilon> 0\) such that for all \(\varepsilon\)-approximate solutions \(f\) of \(P\), we have that \(d(f(x),\sigma_{P(x,z)}) < \frac{\alpha}{3}\) for all \(x \in X\).

Now since \(P\) has approximate roots, we can find an \(\varepsilon\)-approximate solution \(f \in C(X)\) of \(P\); by above we have that \(d(f(x), \sigma_{P(x,z)}) < \frac{\alpha}{3}\) for all \(x \in X\). This means that for each \(x \in X\), there exists some \(\lambda_x \in \sigma_{P(x,z)}\) such that \(|f(x) - \lambda_x| < \frac{\alpha}{3}\); in fact, this \(\lambda_x\) is unique, since if \(\mu_x \in \sigma_{P(x,z)}\) is any other root of \(P(x,z)\), then \[\begin{align} |f(x) - \mu_x| \geq |\lambda_x - \mu_x| - |f(x) - \lambda_x| \geq \delta(x) - \frac{\alpha}{3} \geq 2\alpha - \frac{\alpha}{3} \geq \frac{5}{3}\alpha \end{align}\] Thus we obtain a well-defined function \(g : X \to \mathbb{C}\) given by \(g(x) = \lambda_x\). By definition we have that \(P(x,g(x)) = 0\) for all \(x \in X\), so to show that \(P\) has an exact root, it suffices to show that \(g\) is continuous. To this end, take any \(x_0 \in X\) and assume that \(\varepsilon' > 0\) is given. Since \(f\) is continuous, and by Corollary 8, the function \(X \to \mathbb{C}^n/S_n , x \mapsto \sigma_{P(x,z)}\) is continuous and \(g(x_0) \in \sigma_{P(x_0,z)}\), we can find an open subset \(U\) of \(x_0\) such that whenever \(x \in U\), we have that \(d(g(x_0), \sigma_{P(x,z)}) < \min(\frac{\alpha}{3}, \varepsilon')\) and \(|f(x) - f(x_0)| < \frac{\alpha}{3}\). But this implies that for all \(x \in U\), there is some \(\lambda \in \sigma_{P(x,z)}\) such that \(|\lambda - g(x_0)| < \min(\frac{\alpha}{3}, \varepsilon')\), which implies that \[\begin{align} |\lambda - f(x)| &\leq |\lambda - g(x_0)| + |g(x_0) - f(x_0)| + |f(x_0) - f(x)| \\ &\leq \frac{\alpha}{3} + \frac{\alpha}{3} + \frac{\alpha}{3} = \alpha \end{align}\] The calculation \((\clubsuit)\) above shows that if \(\mu\) is any root of \(P(x,z)\) which is not \(g(x)\), then it must be that \(|f(x) - \mu| \geq \frac{5}{3}\alpha\). But since \(|\lambda - f(x)| \leq \alpha\) and \(\lambda \in \sigma_{P(x,z)}\), this means that \(\lambda = g(x)\). Thus \(|g(x) - g(x_0)| = |\lambda - g(x_0)| < \varepsilon'\). As this holds for all \(x \in U\), this proves that \(g\) is continuous, as desired. ◻

Proof. Obviously if \(P(x,z)\) has an exact root, then it has approximate roots. Conversely assume that \(P(x,z)\) has approximate roots. Then for each integer \(k \geq 1\), we can find a \(\frac{1}{k}\)-approximate root \(f_k \in C(X)\) of \(P(x,z)\). If we can show that \(\{f_k\}_{k = 1}^\infty\) has a convergent subsequence \(\{f_{k_{l}}\}_{l=1}^{\infty}\) converging to some \(f_\infty \in C(X)\), then passing to this subsequence we get \[|P(x, f_\infty(x))| = \lim_{l \to \infty} |P(x, f_{k_l}(x))| \leq \lim_{k \to \infty} \frac{1}{k} = 0\] for each \(x \in X\), showing that \(f_\infty\) is an exact solution. To this end, we will show that \(\{f_k\}_{k = 1}^\infty\) has a convergent subsequence by checking that this sequence is uniformly bounded and equicontinuous, and then applying the Arzelà–Ascoli Theorem.

By Lemma 6, we have that \(\left\Vert f_k \right\Vert_\infty \leq 2 + M\) for all \(k\), showing the sequence is uniformly bounded. For equicontinuity, pick a point \(x_0 \in X\) and \(\varepsilon> 0\). Then we may factor our polynomial at this point \[P(x_0, z) = \prod_{i = 1}^r (z - \lambda_i)^{m_i}\] for some distinct roots \(\lambda_i \in \mathbb{C}\) with respective multiplicities \(m_i\). Let \(\varepsilon_0 := \frac{1}{2}\min(\varepsilon, \min_{i \neq j} |\lambda_i - \lambda_j|)\). By Lemma 7 we can find a neighbourhood \(U_1 \subset X\) of \(x_0\) such that \[\begin{align} \bigcup_{x \in U_1} \{ w \in \mathbb{C} : P(x,w) = 0 \} \subset \bigcup_{i=1}^{r} B_{\frac{\varepsilon_0}{4}}(\lambda_i) \end{align}\] and since \(X\) is locally connected, we can find a connected open neighbourhood \(U\) of \(x_0\) with \(U \subset U_1\). Now by Lemma 9, we can find a positive integer \(k_0\) such that \(d(f_k(x), \sigma_{P(x,z)}) < \frac{\varepsilon_0}{4}\) for all \(k \geq k_0\) and all \(x \in X\). Hence if \(x \in U\) and \(k \geq k_0\), then we can find a root \(w_{k,x}\) of \(P(x,z)\) such that \(|f_k(x) - w_{k,x}| < \frac{\varepsilon_0}{4}\), and an \(i_{k,x} \in \{1, \ldots, r\}\) such that \(w_{k,x} \in B_{\frac{\varepsilon_0}{4}}(\lambda_{i_{k,x}})\). The triangle inequality immediately implies \(|f_k(x) - \lambda_{i_{k,x}}| < \frac{\varepsilon_0}{4} + \frac{\varepsilon_0}{4} = \frac{\varepsilon_0}{2}\) for all \(k\geq k_0\) and all \(x \in U\). It follows that \(f_k(x) \in \bigcup_{i = 1}^r B_{\frac{\varepsilon_0}{2}}(\lambda_i)\) for all \(x \in U\) and \(k \geq k_0\). However, since \(\varepsilon_0 < \min_{i \neq j} |\lambda_i - \lambda_j|\), it follows that the family of disks \(\{B_{\frac{\varepsilon_0}{2}}(\lambda_i)\}_{i=1}^{r}\) are disjoint, so by connectedness of \(U\) it follows that for each \(k \geq k_0\), \(f_k(U)\) must lie in a single such disk; that is, for each \(k \geq k_0\) we can find some \(i_k\) such that \(f_k(U) \subset B_{\frac{\varepsilon_0}{2}}(\lambda_{i_k})\). Therefore for all \(x \in U\) and all \(k \geq k_0\) we have that: \[\begin{align} |f_k(x) - f_k(x_0)| \leq |f_k(x) - \lambda_{i_k}| + |\lambda_{i_k} - f_k(x_0)| < \frac{\varepsilon_0}{2} + \frac{\varepsilon_0}{2} < \varepsilon \end{align}\] Therefore the sequence \(\{f_k\}_{k=k_0}^{\infty}\) is equicontinuous, and since \(f_1, \ldots, f_{k_0}\) are continuous functions, it follows that the sequence \(\{f_k\}_{k=1}^{\infty}\) is equicontinuous as desired. ◻

Proof. Part (i) is clear. For part (ii), let \(M\) be a closed subset of \(\mathbb{C}^n\) and assume that \(f \in \mathcal{A}(X,M)\). Since \(X\) is compact, \(f(X)\) is a compact subset of \(\mathbb{C}^n\) not intersecting \(M\); hence \(\delta := \operatorname{dist}(f(X), M) > 0\). Then if \(\left\Vert g - f \right\Vert < \frac{\delta}{2}\), it follows that for all \(x \in X\) and all \(m \in M\), we have that \[\begin{align} |g(x) - m| \geq |f(x) - m| - |g(x) - f(x)| \geq \delta - \frac{\delta}{2} = \frac{\delta}{2} \end{align}\] Taking the infimum over all \(x \in X\) and all \(m \in M\) we obtain that \(\operatorname{dist}(g(X), M) \geq \frac{\delta}{2} > 0\); therefore \(g(X) \cap M = \emptyset\), so \(g \in \mathcal{A}(X,M)\). Thus \(B_{\frac{\delta}{2}}(f) \subset \mathcal{A}(X,M)\). This proves that \(\mathcal{A}(X,M)\) is open. ◻

Proof.

(i) Take any \(f \in C(X,\mathbb{C}^n)\), and let \(\varepsilon> 0\). Denote by \(f_1,\ldots, f_n\) the components of \(f\). Then \(f_j \in C(X)\) for all \(1 \leq j \leq n\). Since \(\dim X \leq 1\), the invertible elements form a dense subset of \(C(X)\) [13], so we can find non-vanishing functions \(g_1, \ldots, g_n\) in \(C(X)\) such that \(\left\Vert f_j - g_j \right\Vert_{\infty} < \varepsilon\) for each \(1 \leq j \leq n\). Let \(g = (g_1,\ldots, g_n)\). Then \(g \in C(X,\mathbb{C}^n)\), and for each \(x \in X\) we have that \(g_1(x)g_2(x) \cdots g_n(x) \neq 0\), so \(g(x) \notin M\). Therefore \(g \in \mathcal{A}(X,M)\). Moreover, \[\begin{align} \left\Vert f-g \right\Vert = \max_{1 \leq j \leq n} \left\Vert f_j - g_j \right\Vert_{\infty} < \varepsilon \end{align}\] Therefore \(\mathcal{A}(X,M)\) is dense in \(C(X,\mathbb{C}^n)\). Finally, to see that \(\mathcal{A}(X,M)\) is open, just observe that \(M\) is a closed subset of \(\mathbb{C}^n\) and apply Lemma 13.

(ii) Since \(X\) is compact, \(C(X,\mathbb{C}^n)\) has the compact-open topology, so it follows that the the pushforward \(h_* : C(X,\mathbb{C}^n) \to C(X,\mathbb{C}^n)\) is a homeomorphism. By part (i), \(\mathcal{A}(X,M)\) is open and dense in \(C(X,\mathbb{C}^n)\), so it follows that \(h_*(\mathcal{A}(X,M))\) is open and dense in \(C(X,\mathbb{C}^n)\). Then using that \(h\) is bijective, it is easy to see that \(h_*(\mathcal{A}(X,M)) = \mathcal{A}(X,h(M))\), proving (ii).

 ◻

Proof. Here we identify \(\mathbb{C}^n\) with \(\mathbb{R}^{2n}\), and we let \(j\) denote the real dimension of \(N\), which we know is at most \(2n-2\).

Assume first that \(N\) is contained in the graph of a smooth function \(\varphi : \mathbb{R}^j \to \mathbb{R}^{2n - j}\), meaning that \(N \subseteq\{(x, y) \in \mathbb{R}^j \times \mathbb{R}^{2n - j} : y = \varphi(x)\}\) (where by permuting the coordinates we can assume the first \(j\) coordinates are the independent ones). Then letting \(f: \mathbb{R}^{2n} \to \mathbb{R}^{2n}\) be defined by \(f(x, y) = (x, y + \varphi(x))\), which is a diffeomorphism with inverse is given by \((x, y) \mapsto (x, y - \varphi(x))\), we have \[\begin{align} N \subset f(\{(x, y) \in \mathbb{R}^j \times \mathbb{R}^{2n - j} : y = 0\}). \end{align}\] Since the zero set of these last \(2n - j\) coordinates is contained in the zero set of the last two coordinates (which is \(\{z \in \mathbb{C}^n : z_n = 0\}\)), we have completed the proof in this case.

For a general \(N \subseteq\mathbb{R}^{2n}\) we just need to note that \(N\) is locally contained in the graphs of such smooth functions, by the implicit function theorem. Then since \(N\) is second countable, it can be covered by countably many open sets each of which are contained in the graph of such a smooth function. ◻

Proof. Let \(M \subset \mathbb{C}^n\) denote the complex hypersurface \(z_1 z_2 \cdots z_n = 0\). By Lemma 15, we can find a sequence of autodiffeomorphisms \((f_k)_{k=1}^{\infty}\) of \(\mathbb{C}^n\) such that \(N \subset \bigcup_{k=1}^{\infty} f_k(\{z \in \mathbb{C}^n : z_n = 0\})\); hence \(N \subset \bigcup_{k=1}^{\infty} f_k(M)\). In particular, \[\begin{align} \bigcap_{k=1}^{\infty} \mathcal{A}(X,f_k(M)) \subset \mathcal{A}\left(X, \bigcup_{k=1}^{\infty} f_k(M) \right) \subset \mathcal{A}(X,N) \end{align}\] By Lemma 14, each \(\mathcal{A}(X,f_k(M))\) is an open dense subset of the Banach space \(C(X,\mathbb{C}^n)\), so by the Baire Category theorem, \(\bigcap_{k=1}^{\infty} \mathcal{A}(X,f_k(M))\) is a dense subset of \(C(X,\mathbb{C}^n)\). Hence by above, \(\mathcal{A}(X,N)\) is a dense subset of \(C(X,\mathbb{C}^n)\), as desired. ◻

Proof. First, observe that \(V(\Delta)\) is the zero-set of a complex polynomial in \(n\) variables, and hence it has real codimension \(2\) and is a closed subset of \(\mathbb{C}^n\). By [14], the variety \(V(\Delta)\) admits a Whitney stratification \(D = D_2 \cup D_3 \cdots \cup D_n\) where each \(D_j\) consists of those points \((a_0, \ldots, a_{n-1}) \in V(\Delta)\) where the highest multiplicity root of the corresponding polynomial \(a_0 + a_1 z + \cdots + a_{n-1}z^{n-1} + z^n\) is \(j\). Each \(D_j\) is a real smooth submanifold of \(\mathbb{C}^n\) which is a closed subset \(V(\Delta)\) and has codimension at least that of \(V(\Delta)\); hence each has real codimension at least \(2\). It follows by Proposition 16 that \(\mathcal{A}(X, D_j)\) is a dense subset of \(C(X,\mathbb{C}^n)\) for each \(j\), and by Lemma 13, \(\mathcal{A}(X,D_j)\) is also open in \(C(X,\mathbb{C}^n)\). Since the intersection of finitely many open dense sets is open and dense, it follows that \(\mathcal{A}(X,V(\Delta)) = \bigcap_{j=2}^{n} \mathcal{A}(X,D_j)\) is open and dense in \(C(X,\mathbb{C}^n)\), as desired. ◻

Proof. We need to show that \(C(X, B_n)\) is dense in \(C(X, \mathbb{C}^n)\), but the space \(C(X, B_n)\) is exactly \(\mathcal{A}(X, V(\Delta))\), so this follows immediately from Corollary 17 above. ◻

Proof. Let \(M = \max\{ \left\Vert a_0 \right\Vert, \left\Vert a_1 \right\Vert, \ldots, \left\Vert a_{n-1} \right\Vert \}\) and let \(C = \frac{(2+M)^n - 1}{1+M} + 1\). Take a monic polynomial \(Q(x,z) = z^n + b_{n-1}(x)z^{n-1} + \cdots + b_1(x)z + b_0(x)\) with coefficients in \(C(X)\) such that \(\left\Vert a_k - b_k \right\Vert_{\infty} < \varepsilon\). Let \(f \in C(X)\) be an \(\varepsilon\)-approximate solution of \(Q\). Then by Lemma 6, \(\left\Vert f \right\Vert \leq 1 + \varepsilon+ M \leq 2 + M\), so for each \(x \in X\), we have that \[\begin{align} |P(x,f(x))| &\leq |P(x,f(x)) - Q(x,f(x))| + |Q(x,f(x))| \\ &=|(a_{n-1}(x) - b_{n-1}(x))f(x)^{n-1} + \cdots + (a_1(x)-b_1(x))f(x) + (a_0(x)- b_0(x))| \\ & \qquad + |Q(x,f(x))| \\ &\leq \left\Vert a_{n-1} - b_{n-1} \right\Vert\left\Vert f \right\Vert^{n-1} + \cdots + \left\Vert a_1 - b_1 \right\Vert\left\Vert f \right\Vert + \left\Vert a_0 - b_0 \right\Vert + \varepsilon\\ &< \varepsilon(\left\Vert f \right\Vert^{n-1} + \cdots + \left\Vert f \right\Vert + 1) + \varepsilon\\ &\leq \varepsilon\left( (2+M)^{n-1} + \cdots + (2+M) + 1 \right) + \varepsilon\\ &= \varepsilon\left( \frac{(2+M)^n - 1}{(2+M) - 1} + 1 \right) \\ &= C\varepsilon \end{align}\] Since the above holds for all \(x \in X\), it follows that \(\sup_{x \in X} |P(x,f(x))| < C\varepsilon\), as desired. ◻

Proof of Theorem 5. Clearly (1) implies (2), and (2) implies (3) by Theorem 10. Now assume that (3) holds, and take any monic polynomial \(P\) of degree at most \(n\) with coefficients in \(C(X)\). By Lemma 19, we can find a constant \(C > 0\) such that if \(\varepsilon_0 \in (0,1)\) and if \(Q\) is a monic polynomial with coefficients in \(C(X)\) such that \(\left\Vert P - Q \right\Vert < \varepsilon_0\), then every \(\varepsilon_0\)-approximate solution of \(Q\) is a \(C\varepsilon_0\)-approximate solution of \(P\). Now suppose that \(\varepsilon> 0\), and let \(\varepsilon_0 := \min(1, \frac{\varepsilon}{C})\). Since \(\dim X \leq 1\), by Theorem 18, we can find a monic polynomial \(Q\) with coefficients in \(C(X)\) having no repeated roots such that \(\deg Q = \deg P \leq n\) and \(\left\Vert P - Q \right\Vert < \varepsilon_0\). By assumption, \(Q\) has an exact root, say \(f \in C(X)\). Then in particular \(f\) is an \(\varepsilon_0\)-approximate root of \(Q\), so it follows by our choice of \(C\) that \(f\) is a \(C\varepsilon_0\)-approximate solution of \(P\). But \(\varepsilon\geq C \varepsilon_0\), so it follows that \(f\) is an \(\varepsilon\)-approximate solution of \(P\). This proves that (1) holds, as desired. ◻

Proof. Take an inverse system \(\underline{W} = (W_\alpha, w_\alpha, p_{\alpha}^{\alpha'}, A)\) of pointed connected CW complexes and a pointed morphism \(\underline{q}: (X, x_0) \to (\underline{W}, \underline{w})\) so that \((\underline{W}, \underline{q})\) is the pointed shape of \((X, x_0)\).

We start by proving surjectivity. Given a map \(\underline{\varphi}: \underline{\pi}_1(X, x_0) \to G\), we know it is represented by a group homomorphism \(\varphi_\beta : \pi_1(W_\beta, w_\beta) \to G\), for some index \(\beta \in A\). Now applying Proposition 28, this homomorphism is induced by a pointed map \(f_\beta: (W_\beta, w_\beta) \to (Y, y_0)\). But then \(f_\beta \circ q_\beta: (X, x_0) \to (Y, y_0)\) is a map such that \(\underline{\pi}_1(f_\beta \circ q_\beta) = \underline{\varphi}\).

Now for injectivity, consider two pointed maps \(f, g: (X, x_0) \to (Y, y_0)\) such that \(\underline{\pi}_1(f) = \underline{\pi}_1(g)\). Using the definition of the shape of \(X\), there is an index \(\beta \in A\) and pointed maps \(f_\beta, g_\beta: (W_\beta, w_\beta) \to (Y, y_0)\) such that \(f_\beta \circ q_\beta\) is base point homotopic to \(f\) and \(g_\beta \circ q_\beta\) is base point homotopic to \(g\). Since \(\underline{\pi}_1(f) = \underline{\pi}_1(g)\), there is a \(\gamma \geq \beta\) such that the map \(\pi_1(f_\beta \circ p_\beta^\gamma) : \pi_1(W_\gamma. w_\gamma) \to G\) is equal to the map \(\pi_1(g_\beta \circ p_\beta^\gamma)\). But since the pointed maps \(f_\gamma = f_\beta \circ p_\beta^\gamma\) and \(g_\gamma = g_\beta \circ p_\beta^\gamma\) from \((W_\gamma, w_\gamma)\) to \((Y, y_0) \cong K(G, 1)\) induce the same map on \(\pi_1\), they must be base point homotopic by Proposition 28. But given such a homotopy \(f_\gamma \simeq g_\gamma\), we obtain that \[\begin{align} f & \simeq f_\beta \circ q_\beta \simeq f_\beta \circ p_\beta^\gamma \circ q_\gamma = f_\gamma \circ q_\gamma \simeq g_\gamma \circ q_\gamma = g_\beta \circ p_\beta^\gamma \circ q_\gamma \simeq g_\beta \circ q_\beta \simeq g \end{align}\] and so \(f\) and \(g\) are base point homotopic. ◻

Proof. All are a consequence of Lemma 29. Note that \(B_n\) and \(E_n\) are both smooth manifolds and hence they are CW models for the Eilenberg-MacLane spaces \(K(\mathcal{B}_n, 1)\) and \(K(M_n, 1)\) respectively.

To prove (i), we use that the forgetful map \[\underline{\pi}_1 : [(X, x_0), (B, b_0)] \longrightarrow \operatorname{Hom}_{\operatorname{Pro}\mathsf{Grp}}(\underline{\pi}_1(X, x_0), \mathcal{B}_n)\] is surjective to find a polynomial \(P\) that maps to \(\underline{\varphi}\).

For (ii), the “if" direction is immediate, as \(P\) having a solution \(\lambda\) means that the diagram \[\begin{tikzcd} & (E_n, z_i) \\ (X, x_0) & (B_n, b_0) \arrow["P"', from=2-1, to=2-2] \arrow[from=1-2, to=2-2] \arrow["\lambda", from=2-1, to=1-2] \end{tikzcd}\] commutes, which, after applying the functor \(\underline{\pi}_1\), gives the commutative diagram \[\begin{tikzcd} & M_{n,i} \\ \underline{\pi}_1(X, x_0) & \mathcal{B}_n \arrow["\underline{\pi}_1(P)"', from=2-1, to=2-2] \arrow[hook, from=1-2, to=2-2] \arrow["\underline{\pi}_1(\lambda)", from=2-1, to=1-2] \end{tikzcd}\] showing that the image of \(\underline{\pi}_1(P)\) lies in \(M_{n,i}\).

For the converse, assume that the image of \(\underline{\pi}_1(P)\) lies in \(M_{n,i}\), which means that we have a commutative diagram \[\begin{tikzcd} & M_{n,i} \\ \underline{\pi}_1(X, x_0) & \mathcal{B}_n \arrow["\underline{\pi}_1(P)"', from=2-1, to=2-2] \arrow[hook, from=1-2, to=2-2] \arrow["\underline{\varphi}", from=2-1, to=1-2] \end{tikzcd}\] where \(\underline{\varphi}: \underline{\pi}_1(X, x_0) \to M_{n,i}\) is the map given by shrinking the codomain of \(\underline{\pi}_1(P)\). Similarly to part (i), we can use surjectivity of the map in Lemma 29 for the Eilenberg-MacLane space \(E_n = K(M_{n,i}, 1)\) to get a pointed map \(f: (X, x_0) \to (E_n, z_i)\) with \(\underline{\pi}_1(f) = \underline{\varphi}\).

Letting \(\rho: (E_n, z_i) \to (B_n, b_0)\) denote the covering map, we see that \(\rho \circ f: (X, x_0) \to (B_n, b_0)\) is a pointed map such that \[\underline{\pi}_1(\rho \circ f) = \underline{\pi}_1(P)\] Using injectivity of the map in Lemma 29 for the Eilenberg-MacLane space \(B_n\), we see that \(\rho \circ f\) is homotopic to \(P\) (relative to the basepoint). Finally, since \(\rho\) is a covering map, the property of having a lift across \(\rho\) is a homotopy invariant (by lifting the homotopy), so \(\rho \circ f\) having a lift \(f\) implies that \(P\) has a lift \(\lambda\) as required.

Now we consider the statement about complete solvability. The polynomial \(P\) being completely solvable is equivalent to there being \(n\) solutions \(\lambda_i\) on \(X\), so that their values at \(x_0\) are the \(n\) distinct complex numbers \(z_i\); that is, there exist solutions \(\lambda_i: (X, x_0) \to (E_n, z_i)\) for all \(1 \leq i \leq n\). By the first part of (ii) we know that this is equivalent to \(\underline{\pi}_1(P)\) having its image in all of the \(M_{n, i}\). Since the intersection of the \(M_{n, i}\) is \(N_n\) this completes the proof. ◻

Proof. A morphism \(\underline{\varphi}: \underline{G} \to \mathbb{Z}\) is exactly an element of \(A\), and its image lies in \(m\mathbb{Z}\) if and only if there exists some index \(\alpha\) such that the image of a representative \(\varphi_\alpha : G_\alpha \to \mathbb{Z}\) lands in \(m\mathbb{Z}\), if and only if there exists some index \(\alpha\) and homomorphism \(\psi_\alpha : G_\alpha \to \mathbb{Z}\) such that \(\varphi_\alpha = m\psi_\alpha\), if and only if there exists some \(\underline{\psi}: \underline{G} \to \mathbb{Z}\) with \(m \underline{\psi} = \underline{\varphi}\), which is the definition of \(A\) being \(m\)-divisible. ◻

Proof. Consider the subset \(C_m \subseteq B_m\) consisting of polynomials \(z^m - \mu\) for \(\mu \in \mathbb{C}^\times\), which by this parametrization is homeomorphic to \(\mathbb{C}^\times\). Then the preimage of \(C_m\) under our usual covering map \(E_m \to B_m\) is again a copy of \(\mathbb{C}^\times\), and the covering map is given by \(\lambda \mapsto \lambda^m\). Then given such an \(f\) we get a polynomial \(P_f: X \to C_m \subseteq B_m\) by \(P_f(x, z) = z^m - f(x)\) and the question of asking for a solution \(g\) is equivalent to asking for a lift on fundamental groups: \[\begin{tikzcd} & {\mathbb{C}^\times \subseteq E_m} &&& {m\mathbb{Z}} \\ X & {C_m \subseteq B_m} && {\underline{\pi}_1(X,x_0)} & \mathbb{Z} \arrow[from=1-2, to=2-2] \arrow[hook, from=1-5, to=2-5] \arrow[dashed, from=2-1, to=1-2] \arrow["{P_f}"', from=2-1, to=2-2] \arrow[dashed, from=2-4, to=1-5] \arrow["{\underline{\pi}_1(P_f)}"', from=2-4, to=2-5] \end{tikzcd}\] Now, by Proposition 35 since the dual group \[\check{H}^1(X) = \varinjlim \operatorname{Hom}(\underline{\pi}_1(X,x_0), \mathbb{Z})\] is \(m\)-divisible, we get that the image of \(\underline{\pi}_1(P_f)\) lies in \(m\mathbb{Z}\) as required.

Conversely, given a morphism \(\underline{\varphi}: \underline{\pi}_1(X,x_0) \to \mathbb{Z}\) we know it is represented by some \(\varphi_\alpha: \pi_1(W_\alpha, w_\alpha) \to \mathbb{Z}\), where \(\underline{q}: (X, x_0) \to (\underline{W}, \underline{w})\) is the pointed shape of \((X,x_0)\). But \(C_m \cong \mathbb{C}^\times\) is an Eilenberg-MacLane space \(K(\mathbb{Z}, 1)\), so this \(\varphi_\alpha\) is represented by a pointed map \(h_\alpha: (W_\alpha, w_\alpha) \to (\mathbb{C}^\times, 1)\) and \(P_f = h_\alpha \circ q_\alpha: X \to C_m\) is a polynomial \(P_f(z, x) = z^m - f(x)\) with \(\underline{\pi}_1(P_f) = \underline{\varphi}\). By assumption \(f\) has an \(m^{\text{th}}\) root \(g\), which by the above reasoning means that \(\underline{\pi}_1(P_f) = \underline{\varphi}\) lands in \(m\mathbb{Z}\). So \(\check{H}^1(X)\) is \(m\)-divisible by Proposition 35. ◻

Proof. By Corollary 20, (i) is equivalent to the assertion that every monic polynomial with coefficients in \(C(X)\) with no repeated roots can be factored completely. But by [5], since \(\underline{\pi}_1(X)\) is an inverse limit of abelian groups, this is equivalent to (ii). ◻

Proof. Since \(W\) is a finite solvable group, it has a subnormal series \[1 = W_0 \trianglelefteq W_1 \trianglelefteq \cdots \trianglelefteq W_{n - 1} \trianglelefteq W_n = W\] such that \(W_k/W_{k - 1}\) is abelian for each \(1 \leq k\). We shall prove the statement by induction on the length \(n\) of this series. If \(n = 0\) the statement is trivial as \(W\) is already the trivial group.

Now assuming that the statement is true for groups with such a series of length \(n - 1\), take \(W\) to have such a series of length \(n\). Then \(W/W_{n - 1} = W_n/W_{n - 1}\) is an abelian group of exponent (dividing) \(m\). Hence if \[\varphi_\alpha: G_\alpha \to W\] represents \(\underline{\varphi}\), then the composition with the quotient \(W \to W/W_{n - 1}\) factors through \(\mathsf{Ab}(G_\alpha)\), and using the definition of \(\operatorname{Pro}(\mathsf{Ab})(\underline{G})\) being \(m\)-divisible we can find \(\beta \geq \alpha\) so that the image of \(\mathsf{Ab}(G_\beta)\) in \(\mathsf{Ab}(G_\alpha)\) consists only of \(m^{\text{th}}\) powers of elements in \(\mathsf{Ab}(G_\alpha)\). Thus the composition \[\begin{tikzcd} {G_\beta} & {\mathsf{Ab}(G_\beta)} \\ {G_\alpha} & {\mathsf{Ab}(G_\alpha)} & {W/W_{n - 1}} \arrow[from=1-1, to=1-2] \arrow[from=1-1, to=2-1] \arrow[from=1-2, to=2-2] \arrow["0", from=1-2, to=2-3] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \end{tikzcd}\] is trivial. Hence the map from \(G_\beta\) to \(W/W_{n - 1}\) is trivial, and so \(\varphi_\beta: G_\beta \to W\) must actually have its image land in \(W_{n - 1}\). But \(W_{n - 1}\) has a shorter series and therefore the map \(\underline{\varphi}\) is trivial. ◻

Proof. Picking a basepoint \(x_0 \in X\), by Theorem 32 we need to show \(\underline{\pi}_1(X, x_0)\) satisfies \((**_n)\), meaning that any \(\underline{\varphi}: \underline{\pi}_1(X, x_0) \to \mathcal{B}_n\) is such that \(\underline{\psi} = \tau \circ \underline{\varphi}\) is trivial, where \(\tau : \mathcal{B}_n \to S_n\) denotes the canonical map. But since \(n \leq 4\) we have that \(\underline{\psi}: \underline{\pi}_1(X, x_0) \to S_n\) is a morphism into a solvable group of order \(n!\), so By Lemma 40 the abelianization \(\underline{H}_1(X)\) of \(\underline{\pi}_1(X, x_0)\) being \(n!\)-divisible guarantees any morphism into \(S_n\) is trivial. ◻

Proof. For the backwards direction, we can get \(2\)- or \(3\)-divisibility of \(\check{H}^1(X)\) by using the fact that we can find roots of non-vanishing functions (Corollary 36), so let’s look at the forwards direction.

The case \(n = 2\) is simple. As we argued above, \(\check{H}^1(X)\) being \(2\)-divisible ensures that we only need to worry about homomorphisms \(\underline{\varphi}: \underline{\pi}_1(X, x_0) \to \mathcal{B}_2\) whose image lands in \(\mathcal{B}_2'\) which is the trivial group. But these definitely have their image in \(N_2\), so \(\underline{\pi}_1(X, x_0)\) satisfies \((**_2)\).

The case \(n = 3\) is trickier. Consider some \(\underline{\varphi}: \underline{\pi}_1(X, x_0) \to \mathcal{B}_3\), the image of which we may similarly assume lands in \(\mathcal{B}_3'\), as \(\check{H}^1(X)\) is \(6\)-divisible. Since the image of the commutator subgroup \(\mathcal{B}_3'\) under the canonical map \(\tau: \mathcal{B}_3 \to S_3\) lies in the abelian subgroup \(S_3' = A_3 \cong \mathbb{Z}/3\mathbb{Z}\), it follows that \(\tau|_{\mathcal{B}_3'}\) factors through the abelianization of \(\mathcal{B}_3'\). But by the discussion above, \(\mathcal{B}_3'\) is freely generated by two elements, so the abelianization of \(\mathcal{B}_3'\) is \(\mathbb{Z}^2\). In total we get the diagram: \[\begin{tikzcd} & {3\mathbb{Z}^2} \\ {\underline{\pi}_1(X, x_0)} & {\mathbb{Z}^2} & {A_3 \cong \mathbb{Z}/3\mathbb{Z}} \\ & {\mathcal{B}_3'} \arrow[hook, from=1-2, to=2-2] \arrow["0", from=1-2, to=2-3] \arrow[dashed, from=2-1, to=1-2] \arrow["{\underline{\varphi}}", from=2-1, to=3-2] \arrow[from=2-2, to=2-3] \arrow[from=3-2, to=2-2] \arrow["{\tau|_{\mathcal{B}_3'}}"', from=3-2, to=2-3] \end{tikzcd}\] By our assumption that the dual group \[\check{H}^1(X) = \operatorname{Hom}(\underline{\pi}_1(X, x_0), \mathbb{Z})\] is \(3\)-divisible, by Proposition 35 we see that the the map \(\underline{\pi}_1(X, x_0) \to \mathbb{Z}^2\) given by \(\underline{\varphi}\) and then abelianization, has its image in \(3\mathbb{Z}^2\). Therefore \(\tau \circ \underline{\varphi}\) is the trivial map, which in total shows that \(\underline{\pi}_1(X, x_0)\) satisfies \((**_3)\) as required. ◻

Proof. For \(n \geq 5\) we know that the group \(A_n\) is perfect and is generated by two elements \(a, b \in A_n\). In particular we have a surjective morphism \(\psi: \mathbb{Z}* \mathbb{Z}\to A_n\) by sending the free generators \(x, y\) of \(\mathbb{Z}* \mathbb{Z}\) to \(a\) and \(b\) respectively. But now we get a commutative diagram \[\begin{tikzcd} {\mathbb{Z}* \mathbb{Z}} & {(\mathbb{Z}* \mathbb{Z})'} \\ {A_n} & {A_n'} \arrow["\psi"', from=1-1, to=2-1] \arrow[from=1-2, to=1-1] \arrow[from=1-2, to=2-2] \arrow["\operatorname{id}"', from=2-2, to=2-1] \end{tikzcd}\] obtained by restricting to the commutator subgroups, where the bottom map is the identity (as \(A_n'\) is all of \(A_n\)). Therefore the restriction of \(\psi\) to \((\mathbb{Z}* \mathbb{Z})'\) is still surjective, so we can find \(x', y' \in (\mathbb{Z}* \mathbb{Z})'\) such that \(\psi(x') = a\) and \(\psi(y') = b\). We take \(f_n: \mathbb{Z}* \mathbb{Z}\to \mathbb{Z}* \mathbb{Z}\) to be the endomorphism sending \(x\) to \(x'\) and \(y\) to \(y'\). Then by construction \(\psi \circ f_n = \psi\), and hence \(\psi \circ f_n^{\circ m} = \psi\) for any natural number \(m\).

To define \(\varphi\) we lift \(\psi: \mathbb{Z}* \mathbb{Z}\to A_n\) along the natural map \(\tau|_{\mathcal{B}_n'}: \mathcal{B}_n' \to A_n\), which is possible as \(\tau\) (and hence \(\tau|_{\mathcal{B}_n'}\)) is surjective and \(\mathbb{Z}* \mathbb{Z}\) is free. Now we take \(\underline{G^n}\) to be the inverse system where all of the structure maps are \(f_n\), and take the morphism \(\underline{\varphi}: \underline{G^n} \to \mathcal{B}_n' \subseteq\mathcal{B}_n\) given by \(\varphi\) from the first object of the inverse system.

If the image of \(\underline{\varphi}\) were contained in some \(M_{n,i}\), then we should be able to factor \(\varphi\) through \(M_{n,i}\) by going far enough up the inverse system, meaning \(\varphi\circ f_n^{\circ m}\) would have to have its image in \(M_{n,i}\) for some large enough \(m\). However \(M_{n,i}\) is exactly the preimage under \(\tau\) of the subgroup of \(S_n\) consisting of permutations that fix the \(i\)th element, so it is enough to check that \(\tau \circ \varphi\circ f_n^{\circ m}\) never lies in these subgroups. But \[\tau \circ \varphi\circ f_n^{\circ m} = \psi \circ f_n^{\circ m} = \psi\] which has an image of \(A_n\), and no element is fixed by all permutations in this subgroup. Therefore \(\underline{G^n}\) does not satisfy \((*_n)\).

Lastly we need to argue that \(\mathsf{Ab}(\underline{G^n}) = 0\), but this is immediate as \(f_n: \mathbb{Z}* \mathbb{Z}\to \mathbb{Z}* \mathbb{Z}\) has its image landing in the commutator subgroup, so \(f_n\) becomes the zero map after abelianizing and hence the inverse system consists of a sequence of zero maps. ◻

Proof. Let us denote the quotient map by \(f:\operatorname{SL}_2(\mathbb{Z}) \to \operatorname{PSL}_2(\mathbb{Z})\), which has a kernel of \(\{\pm \operatorname{id}\}\). Let’s start with the equivalence between generating a free subgroup of \(\operatorname{SL}_2(\mathbb{Z})\) and \(\operatorname{PSL}_2(\mathbb{Z})\).

If \(U, V\) are free generators of their subgroup \(\left\langle U, V \right\rangle\), then we cannot have \(-\operatorname{id}\) as an element of \(\left\langle U, V \right\rangle\) as this subgroup is torsion-free. Therefore \(\left\langle U, V \right\rangle \cap \ker(f) = \{\operatorname{id}\}\) meaning that \(f\) restricts to be injective on \(\left\langle U, V \right\rangle\) as required. Conversely, if \(f(U), f(V)\) freely generate a subgroup in \(\operatorname{PSL}_2(\mathbb{Z})\) it cannot be that there are any relations between \(U\) and \(V\), as these would give relations between \(f(U)\) and \(f(V)\).

Next we show that under these conditions, the sum of the ranges \(\operatorname{Im}(U - \operatorname{id}) + \operatorname{Im}(V - \operatorname{id}) \subseteq\mathbb{Z}^2\) is a rank \(2\) subgroup. First note that it cannot be that \(U = \operatorname{id}\) or \(V = \operatorname{id}\) if they freely generate a subgroup, therefore \(\operatorname{Im}(U - \operatorname{id})\) and \(\operatorname{Im}(V - \operatorname{id})\) are both at least rank \(1\). Then if \(\operatorname{Im}(U - \operatorname{id}) + \operatorname{Im}(V - \operatorname{id})\) is not rank \(2\), it must be that \(\operatorname{Im}(U - \operatorname{id})\) and \(\operatorname{Im}(V - \operatorname{id})\) are subsets of a common \(\mathbb{Z}w_1\), where by taking out common factors we may assume \(\mathbb{Z}w_1 = (\mathbb{Q}w_1) \cap \mathbb{Z}^2\). In particular \(w_1\) is an eigenvector for both \(U\) and \(V\).

Completing \(w_1\) to a basis \(\{w_1, w_2\}\) for \(\mathbb{Z}^2\), we see that in this basis \(f(U)\) and \(f(V)\) are matrices of the form \[\begin{bmatrix} 1 & * \\ 0 & 1\end{bmatrix}\] inside of \(\operatorname{PSL}_2(\mathbb{Z})\) which commute, and hence cannot freely generate a subgroup. ◻

Proof. Recall that \(\mathcal{B}_4'\) is generated by four elements: \(u, v, a, b\), such that \(a\) and \(b\) freely generate a normal subgroup \(T\) of \(\mathcal{B}_4'\). The elements \(u\) and \(v\) also freely generate a subgroup \(J\). We define the pro-group \(\underline{G}\) by picking an injection \(\mathbb{Z}* \mathbb{Z}\to (\mathbb{Z}* \mathbb{Z})'\) such as the subgroup generated by \(xyx^{-1}y^{-1}\) and \(xy^2x^{-1}y^{-2}\), with which we can recursively define \[\begin{align} u_0 = u \quad & v_0 = v \\ u_n = u_{n - 1}v_{n - 1}u_{n - 1}^{-1}v_{n - 1}^{-1} \quad & v_n = u_{n - 1}v_{n - 1}^2u_{n - 1}^{-1}v_{n - 1}^{-2} \quad \text{for} \;n \geq 1 \end{align}\] to get the subgroups \(J_n = \left\langle u_n, v_n \right\rangle \subseteq J\) and \(G_n = TJ_n \subseteq\mathcal{B}_4'\).

First note that the standard inclusions \(G_n \hookrightarrow \mathcal{B}_4\) induce a morphism \(\underline{\varphi}: \underline{G} \to \mathcal{B}_4\) which does not have its image in one of the \(M_{4, i}\). This is because all of the subgroups \(G_n\) contain \(T = \left\langle a, b \right\rangle = \left\langle \sigma_3 \sigma_1^{-1}, \sigma_2 \sigma_1^{-1}\sigma_3 \sigma_2^{-1} \right\rangle\) whose image under the canonical map \(\mathcal{B}_4 \to S_4\) is the subgroup \(\left\langle (12)(34), (13)(24) \right\rangle\) of \(S_4\). Since no \(i \in \{1,2,3,4\}\) is a common fixed point of these permutations, \(T\) (and hence none of the \(G_n\)) lie in any \(M_{4, i}\), so \(\underline{G}\) doesn’t satisfy \((*_4)\).

We want to understand the groups \(\operatorname{Hom}(G_n, \mathbb{Z})\), which comes down to understanding how \(u_n\) and \(v_n\) act on \(a\) and \(b\). Since \(T\) is a normal subgroup, conjugation by an element of \(J\) descends to an action on the abelianization \(\mathsf{Ab}(T) \cong \mathbb{Z}^2\). By using the images of \(a\) and \(b\) as a basis for \(\mathsf{Ab}(T)\), we get a map \(\alpha: J \to \operatorname{Aut}(\mathsf{Ab}(T)) \cong \operatorname{GL}_2(\mathbb{Z})\) where \(\alpha(x) \in \operatorname{Aut}(\mathsf{Ab}(T))\) is the automorphism defined by \[\alpha(x)(y T') = xyx^{-1} T'.\] Using the relations for \(\mathcal{B}_4'\) we stated above we find \[uau^{-1} = b \mod T'\;\;\;\;\;\; \text{ and } \;\;\; \quad ubu^{-1} = a^{-1}b^3 \mod T'\] so the matrix \(U := \alpha(u)\) in the \(a,b\) basis is given by \[\begin{bmatrix}0 & -1 \\ 1 & 3\end{bmatrix}.\] Similarly for conjugation by \(v\), \[vav^{-1} = a^{-1}b \mod T' \;\;\;\;\;\; \text{ and } \;\;\; \quad vbv^{-1} = a^{-5}b^4 \mod T'\] so the matrix \(V := \alpha(v)\) is given by \[\begin{bmatrix}-1 & -5 \\ 1 & 4\end{bmatrix}.\] Note this computation shows the image of \(\alpha\) lies in \(\operatorname{SL}_2(\mathbb{Z})\). We claim that \(U\) and \(V\) freely generate a subgroup of \(\operatorname{SL}_2(\mathbb{Z})\), and we will use this to prove that \(\varinjlim \operatorname{Hom}(\underline{G}, \mathbb{Z}) = 0\). Note that if \(U_0 = U\) and \(V_0 = V\) freely generate a subgroup, then by induction so do \[\begin{align} U_n & = \alpha(u_n) = U_{n - 1}V_{n - 1}U_{n - 1}^{-1}V_{n - 1}^{-1} \quad \text{ and }\\ V_n & = \alpha(v_n) = U_{n - 1}V_{n - 1}^2U_{n - 1}^{-1}V_{n - 1}^{-2}. \end{align}\] Before proving the claim, let us show that all of the structure maps in \(\operatorname{Hom}(\underline{G}, \mathbb{Z})\) are zero (and hence the limit is zero). By definition the subgroup \(G_n\) is generated by \(T\) and the elements \(u_n\) and \(v_n\). Therefore to understand a structure map \[\begin{tikzcd} {\operatorname{Hom}(G_{n - 1}, \mathbb{Z})} & {\operatorname{Hom}(G_n, \mathbb{Z})} \arrow[from=1-1, to=1-2] \end{tikzcd}\] induced by the inclusion of subgroups, we need to take a morphism \(f: G_{n - 1} \to \mathbb{Z}\) and compute its value on \(T\) and \(u_n\) and \(v_n\). We see that \(f\) automatically vanishes on \(u_n\) and \(v_n\) as they are commutators of elements in \(G_{n - 1}\). Since \(f\) maps into the abelian \(\mathbb{Z}\), its restriction to \(T\) factors through \(\mathsf{Ab}(T) \cong \mathbb{Z}^2\), and to show it is the zero map it is enough to show that \(\tilde{f}: \mathsf{Ab}(T) \cong \mathbb{Z}^2 \to \mathbb{Z}\) has a rank \(2\) subgroup in its kernel. Given any \(x \in J_n\) and \(y \in T\) we have \[\tilde{f}(y T') = \tilde{f}(x T') + \tilde{f}(y T') - \tilde{f}(x T') = \tilde{f}(xyx^{-1} T') = \tilde{f}(\alpha(x)(y T'))\] showing that \(\tilde{f}\) is zero on any elements \((\alpha(x) - \operatorname{id})y\). In particular the images of both \(U_n - \operatorname{id}\) and \(V_n - \operatorname{id}\) are in the kernel of \(\tilde{f}\). Then by the claim we know \(U_n\) and \(V_n\) freely generate a subgroup of \(\operatorname{SL}_2(\mathbb{Z})\), so by Lemma 45 the sum of these images is a rank \(2\) subgroup of \(\mathbb{Z}^2\).

Finally we need to show that \[U = \begin{bmatrix}0 & -1 \\ 1 & 3\end{bmatrix} \quad \text{ and } \quad V = \begin{bmatrix}-1 & -5 \\ 1 & 4\end{bmatrix}\] freely generate a subgroup of \(\operatorname{SL}_2(\mathbb{Z})\), which by Lemma 45 is the same as showing that their images in \(\operatorname{PSL}_2(\mathbb{Z})\) freely generate a subgroup. It is a standard result that \(\operatorname{PSL}_2(\mathbb{Z})\) has the following presentation (see [20]): \[\operatorname{PSL}_2(\mathbb{Z}) = \left\langle S, Q \;| \;S^2 = 1 , \quad Q^3 = 1 \right\rangle \cong \mathbb{Z}/2\mathbb{Z}* \mathbb{Z}/3\mathbb{Z}\] where the matrices \(S\) and \(Q\) are given by \[S = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \quad Q = \begin{bmatrix}1 & -1 \\ 1 & 0\end{bmatrix}\] and our matrices are \(U = S(QS)^3\) and \(V = SQ^{-1}SQU\). We shall compute the subgroup \(\left\langle U, V \right\rangle\) by using covering space theory, which will along the way show that \(\left\langle U, V \right\rangle\) is the commutator subgroup of \(\operatorname{PSL}_2(\mathbb{Z})\).

To start note that when abelianizing \(\operatorname{PSL}_2(\mathbb{Z})\) the element \(U\) is congruent to \(S^4Q^3 = 1\) and \(V\) is congruent to \(S^2U = 1\), so \(\left\langle U, V \right\rangle\) is a subgroup of the commutator subgroup of \(\operatorname{PSL}_2(\mathbb{Z})\). We can form a CW complex \(Y\) with a basepoint \(y_0\), two \(1\)-cells which we label \(S\) and \(Q\), and two \(2\)-cells implementing the relations \(S^2 = 1\) and \(Q^3 = 1\). Thus \(\pi_1(Y, y_0) \cong \operatorname{PSL}_2(\mathbb{Z})\) via the same presentation as above, and we can consider the covering space \((\widetilde{Y}, \widetilde{y_0})\) corresponding to the commutator subgroup of \(\pi_1(Y, y_0)\).

The covering map \((\widetilde{Y}, \widetilde{y_0})\) to \((Y, y_0)\) is a \(6\)-fold covering as the abelianization of \(\operatorname{PSL}_2(\mathbb{Z})\) is \(\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}\), an order \(6\) group. Giving \((\widetilde{Y}, \widetilde{y_0})\) the CW structure induced by the covering map, the \(1\)-skeleton of \((\widetilde{Y}, \widetilde{y_0})\) is as in Figure 1, as the Deck transformations for this space are \(\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}\). Since \(\left\langle U, V \right\rangle\) is a subgroup of the commutator subgroup, the two elements \(U\) and \(V\) still represent loops in \(\widetilde{Y}\) based at \(\widetilde{y_0}\). Collapsing the six \(2\)-cells associated to \(Q^3 = 1\) to two points (one point by identifying \(\widetilde{y_0}\), \(Q\widetilde{y_0}\), \(Q^2\widetilde{y_0}\) and the other by identifying \(S\widetilde{y_0}\), \(QS\widetilde{y_0}\), \(Q^2S\widetilde{y_0}\)) and collapsing the six \(2\)-cells associated to \(S^2 = 1\) to three line segments (one between \(\widetilde{y_0}\) and \(S\widetilde{y_0}\), another between \(Q\widetilde{y_0}\) and \(SQ\widetilde{y_0}\), and the final between \(Q^2\widetilde{y_0}\) and \(SQ^2\widetilde{y_0}\)) we see that \((\widetilde{Y}, \widetilde{y_0})\) is homotopy equivalent to the theta space and hence has \(\pi_1(\widetilde{Y}, \widetilde{y_0}) \cong \mathbb{Z}* \mathbb{Z}\). Moreover, following the loops \(U\) and \(V\) along this homotopy equivalence, we see that they end up as generators for the fundamental group of the theta space (see Figure 2), and hence \(\left\langle U, V \right\rangle = \pi_1(\widetilde{Y}, \widetilde{y_0}) \cong \mathbb{Z}* \mathbb{Z}\) with \(U\) and \(V\) as generators as required. ◻