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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?

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  • $\begingroup$ Does $m=3{{}}$? $\endgroup$ – Lord Shark the Unknown Apr 27 at 17:41
  • $\begingroup$ @LordSharktheUnknown It should have been arbitrary integer $m$. I have corrected ! Thanks! $\endgroup$ – Lev Ban Apr 27 at 17:42
  • $\begingroup$ When you say $m\Bbb Z$, do you mean the subgroup of $\Bbb Z$ generated by $m$? $\endgroup$ – Lord Shark the Unknown Apr 27 at 17:46
  • $\begingroup$ @LordSharktheUnknown Yes, I mean $m\mathbb{Z}=\left< m \right>$ not $\mathbb{Z}/m\mathbb{Z}$. :) $\endgroup$ – Lev Ban Apr 27 at 17:47
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    $\begingroup$ 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 $\mathbb Z$ $\endgroup$ – Max Apr 27 at 17:50
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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 $\Bbb Z/m\Bbb Z$.

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  • $\begingroup$ 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 $\mathbb{Z}$? $\endgroup$ – Lev Ban Apr 27 at 18:03
  • $\begingroup$ I got it! Thanks! $\endgroup$ – Lev Ban Apr 27 at 18:53

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