Homology of $H_1(X_m)$.

Let $$X_m$$ be a space obtained from $$S^1$$ by attaching $$D^2$$ through the map $$f(z)=z^m$$ around the boundary. I have computed the homology group of it by exact sequence

$$\mathbb{Z} \cong H_1(S^1)\xrightarrow{\times m} H_1(X_m) \rightarrow H_1(X_m,S^1)\cong \widetilde{H}_1(X_m/S^1)\cong H_1(X_m/S^1)\cong H_1(S^2)\cong 0.$$

In particular, $$\mathbb{Z}\xrightarrow{\times m} H_1(X_m) \rightarrow 0$$

Thus, the first map is surjective so $$H_1(X_m)\cong m\mathbb{Z}$$.

Am I correct?

• Does $m=3{{}}$? – Lord Shark the Unknown Apr 27 at 17:41
• @LordSharktheUnknown It should have been arbitrary integer $m$. I have corrected ! Thanks! – Lev Ban Apr 27 at 17:42
• When you say $m\Bbb Z$, do you mean the subgroup of $\Bbb Z$ generated by $m$? – Lord Shark the Unknown Apr 27 at 17:46
• @LordSharktheUnknown Yes, I mean $m\mathbb{Z}=\left< m \right>$ not $\mathbb{Z}/m\mathbb{Z}$. :) – Lev Ban Apr 27 at 17:47
• What does $\times m$ mean for a map between two abstract groups ? Perhaps you should look further in the long exact sequence and see that the term on the left of what you wrote also happens to be a $\mathbb Z$ – Max Apr 27 at 17:50

One can represent $$X_m$$ as a CW-complex with one $$0$$-cell, one $$1$$-cell and one $$2$$-cell. One gets a cellular chain complex $$0\to C_2\to C_1\to C_0\to0$$ where each $$C_i\cong\Bbb Z$$. The differential $$C_1\to C_0$$ is zero, so its kernel is $$C_1$$. The differential $$C_2\to C_1$$ takes a generator of $$C_2$$ to $$m$$ times a generator of $$C_1$$. Then $$H_1(X_m)$$ is the homology of this complex at the middle term, which is isomorphic to $$\Bbb Z/m\Bbb Z$$.
• Thank you for detailed answer :) Could I ask why the differential $C_1\rightarrow C_0$ is zero? If it is true, then shouldn't kernel is $\mathbb{Z}$? – Lev Ban Apr 27 at 18:03