I have the following annulus in $\mathbb{C}$:
$U = \left\{z \in \mathbb{C} : 0 < |z|<1 \right\}$ and $V = \left\{z \in \mathbb{C} : 2 < |z|<3 \right\}$
And I have to prove that the induced homomorphism between the fundamental groupus, ie,
$\pi_1(f,z_0): \pi_1(U, z_o) \rightarrow \pi_1(V,f(z_0))$
is the null homormophism. Where f is a holomorphic map $f: U \rightarrow V$
I think that he fundamental groups are the following:
$\pi_1(U, z_o) = \left\{ [\alpha]^k : k \in \mathbb{Z}\right\}$ is the set of loops that go around $0$ k times.
And $\pi_1(V, f(z_o)) = \left\{ [\gamma]^k : k \in \mathbb{Z}\right\}$ is the set of loops that go around $0$ k times.
I can't see the error on defining the induced homomorphism the following way:
$\pi_1(f,z_0): \pi_1(U, z_o) \rightarrow \pi_1(V,f(z_0))$
$\hspace{30mm}$ $[\alpha]^k \rightarrow [\gamma]^k$
Can anyone help me? Thank you in advance.
Assume $f$ nonconstant as otherwise result trivial.
By definition $f$ is bounded hence $0$ is a removable singularity, so $f$ extends to a map from the unit disc $D$ into $V$ (by the open mapping theorem if $f(0)$ is on the boundary of $V$ $f$ is constant and we assumed that is not the case).
Conclude since the fundamental group of the unit disc is $0$ and $\pi_1(f)$ extends to a map $\pi_1(D, z_o) \rightarrow \pi_1(V,f(z_0))$