Theorem  Let Y be a connected open subset of the complex plane, and let B be the closed ball of radius 1 centered at the origin. Suppose B is not entirely contained in Y. Then for any point x in Y−B, the homomorphism from π₁(Y−B,x) to π₁(Y,x) is injective.

Proof   If this homomorphism isn't injective, there's a loop in Y−B that's homotopic to a constant in Y, but not in Y−B. The homotopy h can be considered as a map of the closed unit disc E² to Y.

The loop has a distance of closest approach  d  to B.   Define B' to be the ball of radius R = 1 + d/2. Then the loop is in π₁(Y−B').

Suppose first that Y contains the boundary of B'. Then ε > 0 can be found such that {re^iθ : R−ε < r < R+ε} ⊂ Y. Apply the Seifert-Van Kampen theorem to the open sets U = Y ∩ {re^iθ:   r > R−ε} and V = Y ∩ {re^iθ:  r < R+ε}.   If U were not connected it would be the union of two disjoint open sets U₁ and U₂ with {re^iθ : R−ε < r < R+ε} ⊂ U₁, and U₂ would be disjoint from U₁ ∪ V; so Y would not be connected.  So U is connected.  Similarly V is connected, and because U and V are open sets in the plane, they are path connected.   The maps i₁: π₁ (U ∩ V) → π₁(U) and i₂: π₁ (U ∩ V) → π₁(V) are injections, since π₁ (U ∩ V) is generated by a circular path that has winding number 1 around all points re^iθ with r ≤ R - ε.  Since int(B') ⊄ Y and U ∩ V ⊂ Y,  Y doesn't contain all such points.   The Seifert-Van Kampen theorem says that π₁(U ∪ V) = the free product π₁(U) * π₁(V), mod the relations {i₁(k) = i₂(k), ∀ k in U ∩ V}.   Also π₁(U) = π₁(U) * π₁(U ∩ V)/ {i₁(k) = k}, applying Seifert-Van Kampen to U and U ∩ V.   Since i₂ is injective, π₁(U) * π₁(U ∩ V)/ {i₁(k) = k} is isomorphic to the image of π₁(U) in π₁(U) * π₁(V)/ {i₁(k) = i₂(k)}. So the map π₁(U) → π₁(U ∪ V) is injective.   π₁(Y−B') → π₁(U) is injective, so that proves the loop can't be homotopic to a constant if bdry(B') ⊂ Y.

So assume that bdry(B') ⊄ Y.  Consider the connected component U containing the boundary of E² in h⁻¹ (Y−B').   E²−U is a closed set.  Let {V_i} be the connected components of int(E²−U).   The boundary of a V_i isn't empty since E² is connected.

Lemma: The boundary of a V_i is connected.

Proof: If a boundary point of a V_i has a neighborhood with no intersection with U, the neighborhood is contained in int(E²−U).  So the neighborhood and V_i would be in the same connected component of int(E²−U).  So bdry(V_i) = cl(U) ∩ cl(V_i).   Suppose cl(U) ∩ cl(V_i) isn't connected.  Then it's a union of two disjoint sets, which are closed in cl(U) ∩ cl(V_i), and so also closed in E².   Let b₁ and b₂ be points from the two closed sets.

 Here is an illustration of the proof of the lemma.

By the Urysohn lemma, a function f can be defined on E² with f = 1 on the closed subset of bdry(V_i) containing b₁, and f = −1 on the closed subset of bdry(V_i) containing b₂.

U and the V_i's are path connected because they are connected and open.   Since f = −1 on part of bdry(V_i) = cl(U) ∩ cl(V_i) and f = 1 on the rest of bdry(V_i),   f = 0 on some point in U, and f = 0 on some point in V_i.  

But f isn't 0 on any point of bdry(V_i).   So f ⁻¹(0) is a union of two disjoint closed sets, one contained in V_i and one contained in int(E² − V_i).

So we can use the Urysohn lemma again to define g: cl(E² − V_i) → [0,1]  with g = 1 on the subset of  f ⁻¹(0) in int(E² − V_i), and g = 0 on bdry(V_i). Also define g: cl(V_i) → [−1,0] with g = −1 on the subset of  f ⁻¹(0) in V_i, and g = 0 on bdry(V_i). Since the two definitions agree on a closed set, g is continuous.

Now, define F: E² → C by F(x) = f(x) + ig(x).  F is continuous, and uniformly continuous because it's a function from a compact metric space into a metric space.   F is never 0 on E², so F'(x) = F(x)/|F(x)|  is a (uniformly) continuous map from E² → bdry(E²).

Find a point u₁ in U within a small distance d of b₁, and a point u₂ in U within a distance d of b₂.

Similarly, find a point v₁ in V_i within a distance d of b₁, and a point c₂ in V_i within a distance d of b₂.

Since F'(b₁) = f(b₁) = 1 and F'(b₂) = f(b₂) = −1,   if d is small enough f(u₁), f(v₁) > 0 and f(u₂), f(v₂) < 0, and also
F' on the straight line from u₁ to v₁ doesn't wrap around the origin, and F' on the straight line from u₂ to v₂ doesn't wrap around the origin.

Define a path P₁ in U from u₁ to u₂, and a path P₂ in C_i from v₂ to v₁.

Then there is a loop L from u₁ to u₂ via P₁, from u₂ to v₂ by a straight line, v₂ to v₁ via P₂, from v₁ to u₁ by a straight line.

F'(P₁) is a path from F'(u₁) in the top right quarter-circle through  i  to F'(u₂) in the top left quarter-circle, with the imaginary part never negative.

F' on the straight line from u₂ to v₂ is a path from F'(u₂) in the top left quarter-circle to F'(v₂) in the bottom left quarter-circle, which never has positive real part.

Similarly F'(P₂) is a path from F(v₂) in the bottom left quarter-circle to F'(v₁) in the bottom right quarter-circle, with the imaginary part never positive; and F' on the line from v₁ back to u₁ is a path from F'(v₁) in the bottom right quarter-circle to F'(u₁) in the top right quarter-circle, with non negative real part.

So F' is a map of E² onto bdry(E²), which isn't possible. So bdry(V_i) is connected. ∴

Let {B_j} be the connected components of bdry(U) in E².   The B_j's are contained in h⁻¹ (Y ∩ bdry(B')). Since bdry(V_i) is connected, it is contained in some B_j.  The B_j's are closed, being the components of a closed set.

Some part of the boundary of B' isn't in Y. So, Y ∩ bdry(B')  is a union of disjoint open intervals of bdry(B'). Now define a closed set A_ab for each connected component (a,b) of Y ∩ bdry(B') :

A_ab = cl (∪ B_j such that B_j ⊃ bdry(V_i) for some V_i and h(B_j) is contained in (a,b))

(a,b) is a closed set in the topology of Y, since (a,b) = [a,b] ∩ Y.   And h⁻¹(Y ∩ bdry(B')) is closed in E², so the h⁻¹((a,b))'s are closed sets which don't share connected components of h⁻¹(Y ∩ bdry(B')).   Since h⁻¹((a,b)) is closed, h maps A_ab into (a,b).

Lemma:  Define C_ab = A_ab ∪ V_i such that bdry(V_i) ⊆ some B_j in A_ab.  Then C_ab is closed.
Proof:  Let p be a limit point of C_ab. If every neighborhood of p intersects some B_j in A_ab, then p is in A_ab.  So suppose a neighborhood N of p doesn't intersect A_ab.  Then N intersects a V_i whose boundary is in A_ab. If p is not in C_ab, p is in int(E² − V_i), so a neighborhood N₁ ⊂ N of p is contained in int(E² − V_i).  If N = (N ∩ V_i) ∪ (N ∩ int(E² − V_i)), it would be a union of disjoint open sets, so not connected. So N contains a point of bdry(V_i), so it intersects A_ab.   So C_ab is closed.

By the Tietze extension theorem,  h' can be defined sending C_ab → R, with h' agreeing with h on A_ab.   Since A_ab is compact, h actually sends A_ab into a closed interval [c,d] contained in (a,b).   There is a continuous function j:  R → (a,b) with j(x) = x, ∀ x in [c,d]. Then jh' : C_ab → (a,b) and jh'(x) = h(x), ∀ x ∈ A_ab.

Lemma:   Define h' on E² agreeing with h on cl(U), and h' on the V_i's from the Tietze extension theorem applied to each C_ab, so that  h' doesn't send any of E² into int(B'). Then h' is continuous on E².
Proof:   h' is continuous on U and on the V_i's. Suppose x is in bdry(U), so that h'(x) = h(x) ∈ (a,b) ⊂  Y ∩ bdry(B').   Because  h' = h  on bdry(U), a δ-ball around x can be found small enough so that h'(bdry(U) ∩ δ-ball)  is contained in (a,b).   Suppose a V_i intersects this δ-ball.   If bdry(V_i) ∩ δ-ball = ∅   then δ-ball = (δ-ball ∩ V_i) ∪ (δ-ball ∩ int(E² − V_i)), so it would be disconnected. So bdry(V_i) intersects the δ-ball.   So h'(bdry(V_i)) ⊂ (a,b),  so bdry(V_i) is contained in A_ab and V_i is contained in C_ab. Since h' = h  on cl(U), and h' is defined by the Tietze extension theorem on the closed set C_ab,  and cl(U) ∪ C_ab ⊃ δ-ball, by the glueing lemma  h' is continuous at x.

h' is a homotopy of the loop to a constant and h' doesn't go into int(B').   But then the loop is homotopic to a constant in π₁(Y−B);  remember, B is the ball of radius 1. ∴!

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