If we need a classic single-valued function instead, we have for example:

If \(f\) is another continuous choice of the argument on \(\mathbb{C} \setminus \mathbb{R}_-\) such that \(f(1) = 0,\) the image of \(\mathbb{C} \setminus \mathbb{R}_-\) by the difference \(f - \arg\) is a subset of \(2\pi \mathbb{Z}\) that contains \(0,\) and it’s also path-connected as the image of a path-connected set by a continuous function. Consequently, it is the singleton \(\{0\}\): \(f\) and \(\arg\) are equal. ‌\(\blacksquare\)

The uniqueness of a continuous choice is a consequence of the intermediate value theorem: if we assume that there are two such functions \(\theta_1\) and \(\theta_2\) with the same initial value \(\theta_0,\) as \(\theta_1(0) - \theta_2(0)=0,\) if \(\theta_1(t) - \theta_2(t) \neq 0\) for some \(t\in[0,1],\) then either \(|\theta_1(t) - \theta_2(t)| < \pi,\) or there is a \(\tau \in \left]0,t\right[\) such that \(\theta_1(\tau) - \theta_2(\tau) \neq 0\) and \(|\theta_1(\tau) -\theta_2(\tau)| < \pi.\) In any case, there is a contradiction since all values of the argument differ of a multiple of \(2\pi.\) ‌\(\blacksquare\)

1. The open set \(\Omega = \{(x,y) \in \mathbb{R}^2 \; | \; x < -1 \, \mbox{ or }\, x>1\}\) is not connected but it is simply connected: its complement has a unique component which is unbounded, hence it has no holes.

2. The open set \(\Omega = \mathbb{C} \setminus \overline{\{2^{-n} \, | \, n \in \mathbb{N}\}}\) is multiply connected: its holes are exactly the singletons of its complement.

Intuitively, we should be able to circle around any hole of \(\Omega\) without leaving the set; this idea leads to an alternate characterization of simply connected sets.

1. If \(\gamma\) is a closed path of \(\Omega = \{(x,y) \in \mathbb{R}^2 \; | \; x < -1 \, \mbox{ or }\, x>1\}\) and \(z \in \mathbb{C} \setminus \Omega\), since \(\mathbb{C} \setminus \Omega\) is connected and unbounded, \(z\) belongs to an unbounded component of \(\mathbb{C} \setminus \gamma([0,1]).\) Thus \(\mathrm{ind}(\gamma, z)=0\) for any \(z \in \mathbb{C} \setminus \Omega.\)

2. The open set \(\Omega = \mathbb{C} \setminus \overline{\{2^{-n} \, | \, n \in \mathbb{N}\}}\) is open and multiply connected: for example \(\gamma = 1 + 1/4 [\circlearrowleft]\) is a path of \(\Omega\), \(z=1\) is a point of \(\mathbb{C} \setminus \Omega\) and \(\mathrm{ind}(\gamma, 1) = 1.\)

Conversely, if \(K\) is a hole of \(\Omega,\) then it is a compact set: \(K\) is connected, hence its closure, which is a subset of the closed set \(\mathbb{C} \setminus \Omega,\) is also connected and a superset of \(K;\) as \(K\) is maximal among these sets, \(\overline{K} = K.\) The set \(K\) is therefore closed and bounded, thus is is compact.

Let \(r>0\) such that \(K \subset D(0, r).\) The set \(K\) is a component of the closed set \(A = (\mathbb{C} \setminus \Omega) \cap \overline{D(0,r)}.\) For any point \(a \in A\) on the boundary \(\partial D(0,r)\) of \(D(0,r),\) there is a cover of \(A\) into disjoint open set \(U_a\) and \(V_a\) such that \(a\in U_a\) and \(K \subset V_a.\) Now, the boundary \(\partial D(0,r)\) is compact, thus there is a finite collection of points \(a_1,\) …, \(a_n\) such that \(U = \cup_{i=1}^n U_a\) covers \(A \cap \partial D(0,r).\) The sets \(U\) and \(V = (\cap_{i=1}^n V_a) \cap D(0,r)\) are disjoints open sets that cover \(A\) and \(K \subset V,\) thus the set \(\mathbb{C} \setminus \Omega\) is the disjoint union of the compact set \(L = A \setminus U\) that contains \(K\) and of the closed set \((\mathbb{C} \setminus \Omega) \setminus V\). Therefore, the distance between \(L\) and \(\mathbb{C} \setminus (\Omega \cup L)\) is positive which means that every point of \(L\) is an interior point of \(\Omega \cup L.\) Since every point of \(\Omega\) is also an interior point of \(\Omega \cup L,\) this set is open. ‌\(\blacksquare\)

Conversely, if \(\Omega\) is not simply connected, the set \(\mathbb{C} \setminus \Omega\) has a hole \(K\) which is contained in some compact subset \(L\) of \(\mathbb{C} \setminus \Omega\) such that \(\Omega \cup L\) is open. The distance \(\epsilon\) between \(L\) and \(\mathbb{C} \setminus (\Omega \cup L)\) is positive. Let \(r < \epsilon/\sqrt{2};\) Define for any pair \((k, l)\) of integers the node \(n_{k,l} = (k + i l)r\) and \(S_{k,l}\) as the closed square with vertices \(n_{k,l},\) \(n_{k+1,l},\) \(n_{k+1,l+1}\) and \(n_{k,l+1}.\) The (positively) oriented boundary of the square \(S_{k,l}\) is the polyline \[ [n_{k,l} \to n_{k+1,l} \to n_{k+1,l+1} \to n_{k,l+1} \to n_{k,l}] \] The collection of squares that intersect \(L\) is finite and covers \(L.\) Additionally, all of its squares are included in \(\Omega \cup L.\)

For any square \(S\) in the cover of \(L\) and any interior point \(a\) of \(S\) if \(\gamma\) is the oriented boundary of \(S,\) then \(\mathrm{ind}(\gamma, a) = 1.\) Additionally, \(\mathrm{ind}(\mu, a) = 0\) for the oriented boundary \(\mu\) of any other square in the collection. Consequently, if \(\Gamma\) denotes the collection of oriented line segments that composes the oriented boundaries of all squares of the cover of \(L,\) we have \[ \sum_{\gamma \in \Gamma}\frac{1}{2\pi} [z \mapsto \mathrm{Arg}(z-a)]_{\gamma} = 1. \] Now if the line segment \(\gamma\) belongs to \(\Gamma\) and \(\gamma([0,1]) \cap L \neq \varnothing,\) then \(\gamma^{\leftarrow}\) also belongs to \(\Gamma;\) if we remove all such pairs from \(\Gamma,\) the resulting collection \(\Gamma'\) also satisfies \[ \sum_{\gamma \in \Gamma'}\frac{1}{2\pi} [z \mapsto \mathrm{Arg}(z-a)]_{\gamma} = 1. \] and by construction the image of any \(\gamma\) in \(\Gamma'\) is included in \(\Omega.\) The original collection \(\Gamma\) is balanced: for any square vertice \(n,\) the number of line segments with \(n\) as an initial point and with \(n\) as a terminal point is the same. The collection \(\Gamma'\) has the same property. Consequently, the line segments of \(\Gamma'\) may be assembled in a finite sequence of closed paths \(\gamma_1,\) \(\dots,\) \(\gamma_n\) and \[ \sum_{k=1}^n \mathrm{ind}(\gamma_k, a) = 1. \] Every point of \(L\) is either an interior point of some square of the collection, or the limit of such point; anyway, that means that \[ \forall \, z \in L, \; \sum_{k=1}^n \mathrm{ind}(\gamma_k, z) = 1 \] and thus that there is at least one path \(\gamma_k\) such that \(\mathrm{ind}(\gamma_k, z) \neq 0.\) ‌\(\blacksquare\)

We have for any \(t\in [0,1]\) \[ \mu(t) = \int_{\gamma_t} \frac{dz}{z-a} = \int_0^1 \frac{\gamma'(ts) \times t}{\gamma(t s) - a} ds = \int_0^t \frac{\gamma'(s)}{\gamma(s) - a} ds, \] hence \[ \mu'(t) = \frac{\gamma'(t)}{\gamma(t) - a} \] and the derivative of the quotient \(\phi(t) = {e^{\mu(t)}}/{(\gamma(t)}-a)\) satisfies \[ \phi'(t) = \mu'(t) \phi(t) - \frac{\gamma'(t)}{\gamma(t) - a} \phi(t) = 0 \] which yields the result. ‌\(\blacksquare\)

If additionally \(\gamma\) is a closed path, the equations \[ \gamma(0) = \gamma(1) \; \mbox{ and } \; e^{\mathrm{Re}(\mu(t))} = |\lambda| \times |\gamma(t) - a| \] yield \(e^{\mathrm{Re}(\mu(0))}= e^{\mathrm{Re}(\mu(1))}\) and hence \(\mathrm{Re}(\mu(1)) = \mathrm{Re}(\mu(0)) = 0.\) Thus, \[ \mathrm{ind}(\gamma,a) = \frac{1}{2\pi} \mathrm{Im}(\mu(1)) = \frac{1}{i2\pi} \mu(1), \] which concludes the proof. ‌\(\blacksquare\)