__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
_{1}(Y−B,x)_{1}(Y,x)

__Proof __
If this homomorphism isn't injective, there's a loop in
^{2} to Y.

The loop has a distance of closest approach d to B.
Define B' to be the ball of radius _{1}(Y−B').

Suppose first that Y contains the boundary of B'. Then
^{iθ} : R−ε < r < R+ε}
⊂ Y.^{iθ}: r
> R−ε}^{iθ}: r < R+ε}_{1}
and U_{2} with
{re^{iθ} : R−ε < r < R+ε}
⊂ U_{1}, and U_{2} would be disjoint from
_{1} ∪ V;_{1}: π_{1} (U ∩ V) →
π_{1}(U)_{2}: π_{1} (U ∩ V) →
π_{1}(V)_{1} (U ∩ V) ^{iθ} with r ≤ R - ε._{1}(U ∪ V)_{1}(U) * π_{1}(V),_{1}(k) = i_{2}(k), ∀
k
in U ∩ V}._{1}(U) = π_{1}(U) *
π_{1}(U ∩ V)/ {i_{1}(k) = k},_{2} is injective,
_{1}(U) *
π_{1}(U ∩ V)/ {i_{1}(k) = k}_{1}(U) in
_{1}(U) *
π_{1}(V)/ {i_{1}(k) = i_{2}(k)}._{1}(U) → π_{1}(U ∪ V)_{1}(Y−B') → π_{1}(U)

So
assume that ^{2} in ^{−1} (Y−B').^{2}−U_{i}} be
the connected components of ^{2}−U)._{i} isn't empty since E^{2} 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
^{2}−U)._{i}
would be in the same connected component of
^{2}−U)._{i}) = cl(U) ∩ cl(V_{i})._{i})_{i}),^{2}. Let
b_{1} and b_{2} 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 E^{2} with
_{i})_{1},
and
_{i})_{2}.

U and the V_{i}'s are path connected because they are connected and
open. Since _{i}) = cl(U) ∩ cl(V_{i})_{i})_{i}.

But f isn't 0 on any point of
bdry(V_{i}). So ^{ −1}(0)_{i} and one contained in
^{2} − V_{i}).

So we can use the Urysohn lemma again to define
^{2} − V_{i}) →
[0,1]^{ −1}(0)^{2} − V_{i}),_{i})._{i}) → [−1,0] ^{ −1}(0)_{i}, and
_{i}).

Now, define ^{2} → C by F(x) = f(x) + ig(x).^{2}, so
^{2} → bdry(E^{2}).

Find a point u_{1} in U within a small distance d of b_{1},
and a point u_{2} in U within a distance d of b_{2}.

Similarly, find a point v_{1} in V_{i} within a distance
d of b_{1},
and a point c_{2} in V_{i} within a distance d of
b_{2}.

Since _{1}) = f(b_{1}) = 1_{2}) = f(b_{2}) = −1,_{1}), f(v_{1}) > 0_{2}), f(v_{2}) < 0,

F' on the straight line from u_{1} to v_{1} doesn't
wrap around the origin, and F' on the straight line
from u_{2} to v_{2} doesn't wrap around the origin.

Define a path P_{1} in U from u_{1} to u_{2},
and a path P_{2} in C_{i} from v_{2} to
v_{1}.

Then there is a loop L from u_{1} to u_{2} via P_{1},
from u_{2} to v_{2}
by a straight line, v_{2} to v_{1} via P_{2}, from
v_{1} to u_{1} by a straight line.

_{1})_{1})_{2})

F' on the straight line from u_{2} to v_{2} is a
path from _{2})_{2})

Similarly _{2})_{2})_{1})_{1} back to u_{1} is a path from
_{1}) _{1})

So F' is a map of E^{2} onto ^{2}),_{i})

Let _{j}}^{2}.
The B_{j}'s are contained in ^{−1}
(Y ∩ bdry(B'))._{i})_{j}. The _{j}'s

Some part of the boundary of B' isn't in Y.
So, _{ab} for each connected component

_{ab} = cl (∪ B_{j} _{j} ⊃ bdry(V_{i})_{i} and _{j})

^{−1}(Y ∩ bdry(B'))^{2},
so the ^{−1}((a,b))'s^{−1}(Y ∩ bdry(B')).^{−1}((a,b))_{ab} into (a,b).

__Lemma:__ Define
_{ab} = A_{ab} ∪ V_{i} such that
bdry(V_{i}) ⊆ some B_{j} in
A_{ab}._{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
^{2} − V_{i})_{1} ⊂ N of p^{2} − V_{i})_{i}) ∪
(N ∩ int(E^{2} − V_{i})),_{i}),_{ab}.
So C_{ab} is closed.

By the
Tietze extension theorem, h' can be defined sending
_{ab} → R,_{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
_{ab} → (a,b) _{ab}.

__Lemma:__ Define h' on E^{2} agreeing with h on cl(U),
and h' on the _{i}'s_{ab}, so that h'
doesn't send any of E^{2} into int(B').
Then h' is continuous on E^{2}.

__Proof:__
h' is continuous on U and on the V_{i}'s. Suppose x is in
_{i} intersects this _{i}) ∩ δ-ball = ∅_{i}) ∪ (δ-ball
∩ int(E^{2} − V_{i})),_{i})_{i})) ⊂ (a,b),_{i}) is contained in A_{ab} and
V_{i} is contained in C_{ab}. Since
_{ab},_{ab} ⊃ δ-ball,

h' is a homotopy of the loop to a constant and h' doesn't go into
_{1}(Y−B);

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