1999, polished up 2007.

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 : R−ε < r < R+ε} ⊂ Y. Apply the Seifert-Van Kampen theorem to the open sets U = Y ∩ {re:   r > R−ε} and V = Y ∩ {re:  r < R+ε}.   If U were not connected it would be the union of two disjoint open sets U1 and U2 with {re : 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 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 {Vi} be the connected components of int(E2−U).   The boundary of a Vi isn't empty since E2 is connected.

Lemma: The boundary of a Vi is connected.

Proof: If a boundary point of a Vi has a neighborhood with no intersection with U, the neighborhood is contained in int(E2−U).  So the neighborhood and Vi would be in the same connected component of int(E2−U).  So bdry(Vi) = cl(U) ∩ cl(Vi).   Suppose cl(U) ∩ cl(Vi) isn't connected.  Then it's a union of two disjoint sets, which are closed in cl(U) ∩ cl(Vi), 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(Vi) containing b1, and f = −1 on the closed subset of bdry(Vi) containing b2.

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

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

So we can use the Urysohn lemma again to define g: cl(E2 − Vi) → [0,1]  with g = 1 on the subset of  f −1(0) in int(E2 − Vi), and g = 0 on bdry(Vi). Also define g: cl(Vi) → [−1,0] with g = −1 on the subset of  f −1(0) in Vi, and g = 0 on bdry(Vi). 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 Vi within a distance d of b1, and a point c2 in Vi 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 Ci 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(Vi) is connected. ∴

Let {Bj} be the connected components of bdry(U) in E2.   The Bj's are contained in h−1 (Y ∩ bdry(B')). Since bdry(Vi) is connected, it is contained in some Bj.  The Bj'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 Aab for each connected component (a,b) of Y ∩ bdry(B') :

Aab = cl (∪ Bj such that Bj ⊃ bdry(Vi) for some Vi and h(Bj) 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 Aab into (a,b).

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

By the Tietze extension theorem,  h' can be defined sending Cab → R, with h' agreeing with h on Aab.   Since Aab is compact, h actually sends Aab 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' : Cab → (a,b) and jh'(x) = h(x), ∀ x ∈ Aab.

Lemma:   Define h' on E2 agreeing with h on cl(U), and h' on the Vi's from the Tietze extension theorem applied to each Cab, 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 Vi'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 Vi intersects this δ-ball.   If bdry(Vi) ∩ δ-ball = ∅   then δ-ball = (δ-ball ∩ Vi) ∪ (δ-ball ∩ int(E2 − Vi)), so it would be disconnected. So bdry(Vi) intersects the δ-ball.   So h'(bdry(Vi)) ⊂ (a,b),  so bdry(Vi) is contained in Aab and Vi is contained in Cab. Since h' = h  on cl(U), and h' is defined by the Tietze extension theorem on the closed set Cab,  and cl(U) ∪ Cab ⊃ 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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