Homology of $H_1(X_m)$.

Question

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?

@LordSharktheUnknown It should have been arbitrary integer $m$. I have corrected ! Thanks! — Lev Bahn, Apr 27 '19 at 17:42
When you say $m\Bbb Z$, do you mean the subgroup of $\Bbb Z$ generated by $m$? — Angina Seng, Apr 27 '19 at 17:46
@LordSharktheUnknown Yes, I mean $m\mathbb{Z}=\left< m \right>$ not $\mathbb{Z}/m\mathbb{Z}$. :) — Lev Bahn, Apr 27 '19 at 17:47
So then $m\Bbb Z\cong\Bbb Z$ so you are saying $H_1(X_m)\cong\Bbb Z$? — Angina Seng, Apr 27 '19 at 17:48
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 $\mathbb Z$ — Maxime Ramzi, Apr 27 '19 at 17:50
@LordSharktheUnknown I think so! But then my question is if $H_1(X_m)\cong \mathbb{Z}$... — Lev Bahn, Apr 27 '19 at 17:50

Angina Seng · Accepted Answer · 2019-04-27T18:03:32.737

3

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 $\Bbb Z/m\Bbb Z$.

edited Apr 27 '19 at 18:03

answered Apr 27 '19 at 17:57

Angina Seng

158,341

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 Bahn Apr 27 '19 at 18:03
I got it! Thanks! – Lev Bahn Apr 27 '19 at 18:53

Homology of $H_1(X_m)$.

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