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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.

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 $frac{a}{lvert a rvert} = e^{itau}$ with $tau in [0,2pi)$. A homotopy $H : S^1 times [0,1] to S^1$ is defined by
$$H(z,t) = e^{ittau}z^m .$$
Then $H(z,0) = mu(z), H(z,1) = varphi(z)$. Hence $mu$ and $varphi$ are homotopic and we conclude $varphi_* = mu_*$.