**Proof plan.** We verify each property in order. Convergence (Claim 1) follows from the [integrability](/page/Integral) of $|\zeta - x_k|^{\alpha_k - 1}$ when $\alpha_k - 1 > -1$. Holomorphicity and local injectivity (Claim 2) are immediate from the fact that the integrand is holomorphic and nonvanishing on $\mathbb{H}$. The [boundary](/page/Boundary) correspondence (Claim 3) uses an argument-of-the-[derivative](/page/Derivative) analysis along the real axis. Global injectivity (Claim 4) is the most delicate step and follows from the boundary correspondence combined with [the argument principle](/theorems/356). Surjectivity (Claim 5) follows from [topological](/page/Topology) degree theory for proper holomorphic maps. **Step 1: Convergence and path-independence of the integral.** [claim:Convergence And Path Independence Of The Schwarz Christoffel Integral] For every $z \in \overline{\mathbb{H}} \setminus \{x_1, \ldots, x_n\}$ and every piecewise smooth path $\gamma$ from $z_0$ to $z$ lying in $\overline{\mathbb{H}} \setminus \{x_1, \ldots, x_n\}$, the integral $\int_\gamma \prod_{k} (\zeta - x_k)^{\alpha_k - 1} \, d\zeta$ converges absolutely. Furthermore, for any path approaching a finite prevertex $x_j$, the improper integral converges. Finally, the value of the integral depends only on the endpoints $z_0$ and $z$, not on the choice of path within $\mathbb{H}$. [/claim] [proof] The integrand is continuous and bounded on compact subsets of $\overline{\mathbb{H}} \setminus \{x_1, \ldots, x_n\}$, so the only convergence issue arises near the prevertices and at infinity. *Near a finite prevertex $x_j$.* For $\zeta$ near $x_j$, the singular factor satisfies $|(\zeta - x_j)^{\alpha_j - 1}| = |\zeta - x_j|^{\alpha_j - 1}$, while the remaining factors $\prod_{k \neq j} |\zeta - x_k|^{\alpha_k - 1}$ are bounded above and below by positive constants in a neighbourhood of $x_j$. Therefore the integrand is $O(|\zeta - x_j|^{\alpha_j - 1})$ near $x_j$. Since $\alpha_j \in (0, 2)$, the exponent $\alpha_j - 1 \in (-1, 1)$, and in particular $\alpha_j - 1 > -1$. The integral $\int_0^\epsilon r^{\alpha_j - 1} \, dr = \epsilon^{\alpha_j} / \alpha_j < \infty$ converges, so the singularity is integrable. *At infinity.* When all prevertices are finite, the integrand satisfies $|\prod_k (\zeta - x_k)^{\alpha_k - 1}| \sim |\zeta|^{\sum_k (\alpha_k - 1)} = |\zeta|^{-2}$ as $|\zeta| \to \infty$ (by the angle-sum identity). Since $|\zeta|^{-2}$ is integrable near infinity in $\mathbb{C}$, any path to $\infty$ gives a convergent integral. If one prevertex is at infinity, the corresponding factor is absent, and the decay is $|\zeta|^{-2 + (1 - \alpha_m)} = |\zeta|^{-(1 + \alpha_m)}$ with $1 + \alpha_m > 1$, which is still integrable. *Path-independence.* The integrand $F(\zeta) = \prod_k (\zeta - x_k)^{\alpha_k - 1}$ is holomorphic on $\mathbb{H}$: each factor $(\zeta - x_k)^{\alpha_k - 1}$ is defined using the branch with $\arg(\zeta - x_k) \in (0, \pi)$, which is a well-defined holomorphic function on $\mathbb{H}$ because the [branch cut](/page/Branch%20Cuts) (the ray $(-\infty, x_k]$) lies on $\mathbb{R}$ and does not enter $\mathbb{H}$. Since $\mathbb{H}$ is simply connected, [Cauchy's Theorem for Simply Connected Domains](/theorems/344) guarantees that $\int_\gamma F(\zeta) \, d\zeta = 0$ for every closed piecewise smooth loop $\gamma$ in $\mathbb{H}$. It follows that $\int_{\gamma_1} F \, d\zeta = \int_{\gamma_2} F \, d\zeta$ for any two paths $\gamma_1, \gamma_2$ in $\mathbb{H}$ sharing the same endpoints, so the integral depends only on $z_0$ and $z$. [/proof] **Step 2: Holomorphicity and local injectivity.** [claim:Holomorphicity And Local Injectivity] The function $f$ is holomorphic on $\mathbb{H}$, and $f'(z) \neq 0$ for all $z \in \mathbb{H}$. [/claim] [proof] Since the integrand $F(\zeta) = c_1 \prod_k (\zeta - x_k)^{\alpha_k - 1}$ is holomorphic on $\mathbb{H}$ (the branch points $x_k$ lie on $\mathbb{R}$, which is outside $\mathbb{H}$), the antiderivative $f(z) = \int_{z_0}^z F(\zeta) \, d\zeta + c_2$ is holomorphic on $\mathbb{H}$ with derivative $f'(z) = F(z)$. For $z \in \mathbb{H}$, each factor $|z - x_k| > 0$ (since $\operatorname{Im}(z) > 0$ and $x_k \in \mathbb{R}$), and $c_1 \neq 0$ by hypothesis. Therefore $f'(z) = c_1 \prod_k (z - x_k)^{\alpha_k - 1} \neq 0$ for all $z \in \mathbb{H}$, and $f$ is a local biholomorphism (hence locally conformal) by the [inverse function theorem](/page/Inverse%20Function%20Theorem) for holomorphic maps. [/proof] **Step 3: Boundary correspondence and edge structure.** [claim:Boundary Correspondence] The map $f$ extends [continuously](/page/Continuity) to $\mathbb{R} \setminus \{x_1, \ldots, x_n\}$, and on each interval $(x_k, x_{k+1})$ the image $f(t)$ traces a straight line segment. At each prevertex $x_k$, the images of the adjacent intervals meet at an angle of $\alpha_k \pi$. [/claim] [proof] *Straight edges.* For $t \in (x_k, x_{k+1})$ (all prevertices finite), each factor $(t - x_j)$ is real: positive for $j \leq k$ and negative for $j > k$. The argument of $f'(t)$ is therefore \begin{align*} \arg f'(t) = \arg c_1 + \sum_{j=k+1}^{n} (\alpha_j - 1) \pi, \end{align*} which is independent of $t$. Since $f'(t)$ has constant argument and $f(t) = \int_{z_0}^t f'(s) \, ds + c_2$, the image $f(t)$ moves in a fixed direction as $t$ traverses $(x_k, x_{k+1})$, producing a straight line segment. *Angle at prevertices.* As $t$ increases through $x_k$, the factor $(t - x_k)^{\alpha_k - 1}$ transitions from argument $(\alpha_k - 1)\pi$ (for $t < x_k$) to argument $0$ (for $t > x_k$). The resulting jump in $\arg f'$ is $(\alpha_k - 1)\pi$, exactly as in the proof of the [Schwarz–Christoffel Half-Plane Formula](/theorems/683). The exterior angle at vertex $w_k$ is therefore $(1 - \alpha_k)\pi$, so the interior angle is $\pi - (1 - \alpha_k)\pi = \alpha_k \pi$ as required. *Continuity at prevertices.* The function $f$ itself (not $f'$) extends continuously to $x_k$: by the convergence established in Claim 1, the [limit](/page/Limit) $f(x_k) = \lim_{t \to x_k} f(t)$ exists and is finite, since the integral $\int_{z_0}^{x_k}$ converges. [/proof] **Step 4: Global injectivity.** [claim:Global Injectivity On The Upper Half Plane] The map $f$ is injective on $\mathbb{H}$. [/claim] [proof] We first establish injectivity on the boundary, then lift to the interior. *Boundary injectivity.* On each interval $(x_k, x_{k+1})$, the derivative $f'(t)$ is real and nonzero (since $f'(t) = c_1 \prod_j |t - x_j|^{\alpha_j - 1} e^{i\theta_k}$ for a constant phase $\theta_k$), so $|f'(t)| > 0$ and $f$ is strictly monotone along each edge. The image of each interval is therefore a non-degenerate line segment. At each prevertex $x_k$, the direction changes by the exterior angle, and the angle-sum identity $\sum_k (1 - \alpha_k) = 2$ ensures that the total turning is $2\pi$, so the boundary closes into a simple polygon. Since the edges are straight and the total turning is exactly $2\pi$, the boundary curve $f|_{\mathbb{R}}$ is a simple (non-self-intersecting) closed curve. *Interior injectivity.* Since $f'(z) \neq 0$ on $\mathbb{H}$, the map $f$ is a local homeomorphism. The image of $\mathbb{R} \cup \{\infty\}$ is a simple closed curve $\partial P$ (established above). For any $w_0$ in the interior of this curve, the winding number $n(f \circ \gamma, w_0) = 1$ for a large semicircular contour $\gamma$ in $\overline{\mathbb{H}}$ (by the argument principle and the fact that the boundary has winding number $\pm 1$). By the argument principle for holomorphic [functions](/page/Function), $f$ takes the value $w_0$ exactly once in $\mathbb{H}$ (counted with multiplicity; since $f' \neq 0$, every preimage is simple). Therefore $f$ is injective on $\mathbb{H}$. [/proof] **Step 5: Surjectivity.** By Claim 4, the map $f : \mathbb{H} \to \mathbb{C}$ is an injective holomorphic function whose boundary values trace the simple closed polygon $\partial P$. For any point $w_0$ in the interior of $P$, the argument principle (used in Claim 4) shows that $f(z) = w_0$ has exactly one solution in $\mathbb{H}$. Therefore $f(\mathbb{H})$ contains the entire interior of $P$. Conversely, $f(\mathbb{H})$ is contained in the interior of $P$: since $f$ is an open map (by the [open mapping theorem for holomorphic functions](/theorems/358)) and $f(\mathbb{R})= \partial P$, the image of $\mathbb{H}$ lies on one side of $\partial P$. The orientation convention (counterclockwise listing of vertices, interior-on-left) ensures that this side is the interior. Therefore $f : \mathbb{H} \to \operatorname{int}(P)$ is a conformal bijection.