Induced homomorphism $p_*: H_1(D_0 - z_0)\to H_1(\mathbb{C}-0)$ by $a(z-z_0)^m$

Question

Let $D_0 \subseteq \mathbb{R}^2$ be a closed disc with center at $z_0$ and consider the function $p: D_0 \to \mathbb{C}$ defined by $p(z):=a(z-z_0)^m$ for $z \in D_0$ with $a \neq 0$. How can I show that the induced homomorphism $p_*: H_1(D_0 - z_0) \to H_1(\mathbb{C}-0)$ is multiplication by $m$?. I know that $H_1(D_0 - z_0)$ and $H_1(\mathbb{C}-0)$ are infinite cyclic groups and that the homorphism induced by $z^m$ in $H_1(S^1) \to H_1(S^1) $ is multiplication by $m$. I think that I should work with the isomorphism between $H_1(\mathbb{C}-0)$ and $H_1(S^1)$ and the isomorphism of $H_1(D_0 - z_0)$ with $H_1(\partial D_0)$.

Thanks for any help.

Answer 1

Certainly we have $H_1(D_0 \setminus z_0) \approx \mathbb{Z}$ and $H_1(\mathbb{C} \setminus 0) \approx \mathbb{Z}$. However, to see what $p_*$ looks like we must specify generators $a$ of $H_1(D_0 \setminus z_0)$ and $b$ of $H_1(\mathbb{C} \setminus 0)$. There are two possible choices for both homology groups. This is equivalent to specifying isomorphisms $f : H_1(S^1) \to H_1(D_0 \setminus z_0)$ and $g : H_1(S^1) \to H_1(\mathbb{C} \setminus 0)$. This is done in the following obvious way:

(1) There exists a strong deformation retraction $r : \mathbb{C} \setminus 0 \to S^1, r(z) = z/\lVert z \rVert$. Hence the inclusion $i : S^1 \to \mathbb{C} \setminus 0$ is a homotopy equivalence and induces the isomorphism $g = i_*$. Its inverse is $g^{-1} = r_*$.

(2) $\partial D_0$ is a strong deformation retract of $D_0 \setminus z_0$. But there is a canonical homeomorphism $h : S^1 \to \partial D_0, h(z) = z_0 + dz$, where $d$ is the radius of $D_0$. If $j : \partial D_0 \to D_0$ denotes inclusion, we take $f = (j \circ h)_*$.

You know that the map $\mu : S^1 \to S^1, \mu(z) = z^m,$ induces $\mu_*$ = multiplication by $m$.

Now we can check that $g^{-1} \circ p_* \circ f$ is multiplication by $m$. We have $$g^{-1} \circ p_* \circ f = r_* \circ p_* \circ (j \circ h)_* = (r \circ p \circ j \circ h)_*$$ and $$(r \circ p \circ j \circ h)(z) = r(p(z_0+dz)) = r(adz^m) = \frac{a}{\lvert a \rvert} z^m .$$ Let $\varphi = r \circ p \circ j \circ h$ and write $\frac{a}{\lvert a \rvert} = e^{i\tau}$ with $\tau \in [0,2\pi)$. A homotopy $H : S^1 \times [0,1] \to S^1$ is defined by $$H(z,t) = e^{it\tau}z^m .$$ Then $H(z,0) = \mu(z), H(z,1) = \varphi(z)$. Hence $\mu$ and $\varphi$ are homotopic and we conclude $\varphi_* = \mu_*$.