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This exercise is inspired from an exercise found in Hatcher's book, p. 156, 13:

Compute the homology of the space $X$ obtained from the $1$-sphere $\mathbb{S}^1$ by attaching two $2$-cells via attaching maps $\mathbb{S}^1 \to \mathbb{S}^1$ of degree $2$ and $3$.

My result is: $$\begin{align*}H_k(X) = \begin{cases} \mathbb{Z} & k = 0,\\ \mathbb{Z}/(2\mathbb{Z} \oplus 3\mathbb{Z}) & k = 1,\\ 0 & k >1 \end{cases}\end{align*}$$

is this correct?

Edit. Giving it a second look, writing $\mathbb{Z}/(2\mathbb{Z} \oplus 3\mathbb{Z})$ is blatantly stupid. Now I understand the "matrix notation" provided in the answer and can apply it to the cellular chain complex of the space $X$: $$0 \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} \to 0$$ If we write $s$ for the $2$-cell with degree $2$ and $t$ for the $2$-cell with degree $3$, the cellular boundary formula yields $$\partial s = 2a \qquad \text{and} \qquad \partial t = 3 a$$ where $a$ denotes the $1$-cell of the cell-complex.

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    $\begingroup$ $2\Bbb Z\oplus 3\Bbb Z$ is not a subgroup of $\Bbb Z$. $\endgroup$ Jan 22, 2018 at 19:55
  • $\begingroup$ @LordSharktheUnknown well...I posted my idea too early.. $\endgroup$ Jan 22, 2018 at 19:56
  • $\begingroup$ It is not that your idea was posted too early: is that what you wrote does not make sense (My guess is, you wanted to write something like $Z/(2Z+3Z)$, with a sum and not a direct sum) $\endgroup$ Jan 22, 2018 at 20:00
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    $\begingroup$ And, by the way, you did not post any idea. You just wrote down the result. $\endgroup$ Jan 22, 2018 at 20:00
  • $\begingroup$ @MarianoSuárez-Álvarez with idea I meant solution...sorry. $\endgroup$ Jan 22, 2018 at 20:01

1 Answer 1

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The cellular chain group is given by $C_2=\Bbb Z^2$, $C_1=\Bbb Z$ and $C_0=\Bbb Z$. The map from $C_2$ to $C_1$ is given by the matrix $(2\ 3)$ and that from $C_1$ to $C_0$ is zero. The $C_2\to C_1$ map is surjective, but has kernel $\cong \Bbb Z$, so $H_2(X)\cong \Bbb Z$ while $H_1(X)=0$.

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  • $\begingroup$ Could you let me know why the map $C_2\rightarrow C_1$ is surjective? $\endgroup$
    – Lev Bahn
    Apr 27, 2019 at 20:29
  • $\begingroup$ $-(2)+3=1$ @LevBan $\endgroup$ Apr 28, 2019 at 6:39

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