Let $X$ be the $2$-complex obtained from $S^1$ with its usual cell structure by attaching two $2$-cells by maps of degrees $2$ and $3$, respectively.

Hatcher ask us to compute the homology group of $X/A$ where $A$ is any subcomplex of $X$. The subcomplexes consist of $e^0, S^1, S^1 \cup_2 e^2_1, S^1 \cup_3 e^2_2$ and $X$. Using the cellular boundary formula, I could compute their homology groups, namely, \begin{align*} \tilde{H}_n(X) \cong \begin{cases} \mathbb{Z} & n=2 \\ 0 & \text{otherwise}, \end{cases} \end{align*}

\begin{align*} \tilde{H}_n(S^1 \cup_2 e^2_1) \cong \begin{cases} \mathbb{Z}_2 & n=1 \\ 0 & \text{otherwise}, \end{cases} \end{align*} and \begin{align*} \tilde{H}_n(S^1 \cup_3 e^2_2) \cong \begin{cases} \mathbb{Z}_3 & n=1 \\ 0 & \text{otherwise}. \end{cases} \end{align*} I am not able to compute all homology groups of $\tilde{H}_\ast(X/S^1)$ and $\tilde{H}_\ast(X/S^1 \cup_{2,3} e^2_{1,2})$ for those three subcomplexes. Here is what I have tried so far:

I used the long exact sequence in reduced homology for the pair $(X,A)$ with $A$ the subcomplex. For example, the pair $(X,S^1 \cup_2 e^2_1)$ gives \begin{equation} 0 \longrightarrow \mathbb{Z} \longrightarrow H_2(X/(S^1 \cup_2 e^2_1)) \longrightarrow \mathbb{Z}_2 \longrightarrow 0 \longrightarrow H_1(X/(S^1 \cup_2 e^2_1)) \longrightarrow 0. \end{equation} Hence, we deduce that $H_1(X/(S^1 \cup_2 e^2_1))$ is trivial by exactness of the sequence. How can go I from there? I do not understand how to compute the induced inclusion $j_\ast : H_2(X) \rightarrow H_2(X/(S^1 \cup_2 e^2_1))$ and the connecting homomorphism $\partial_\ast : H_2(X/(S^1 \cup_2 e^2_1)) \rightarrow H_1(S^1 \cup_2 e^2_1)$.


1 Answer 1


I will use $\cong$ to indicate homeomorphic. Note that $$\frac{X}{\Bbb S^1}=\frac{X^{(2)}}{X^{(1)}}\cong \bigvee_{2-\text{cells}}\Bbb S^2= \Bbb S^2\lor \Bbb S^2,$$ $$\frac{X}{\Bbb S^1\sqcup_2 e^2_1}\cong \frac{\Bbb D^2}{\Bbb S^1}\cong\Bbb S^2,\text{ and }\frac{X}{\Bbb S^1\sqcup_3 e^2_2}\cong \frac{\Bbb D^2}{\Bbb S^1}\cong\Bbb S^2.$$

Now, you can compute all $\widetilde H_*(X/A)$, keep in mind that $\widetilde H_*(\Bbb S^2\lor \Bbb S^2)=\widetilde H_*(\Bbb S^2)\oplus \widetilde H_*(\Bbb S^2).$

The quotient of $n$-th skeleton by $(n-1)$-th skeleton is a wedge of $\#n\text{-cells}$ many $n$-sphers.

Note that we are also using following general fact: Let $X=\sqcup_\alpha\Bbb D^n$ and $A=\sqcup_\alpha\Bbb S^{n-1}$ and $Y$ be arbitrary space with a map $f:A\to Y$. Now, $Y$ is a closed subspace of $Y\sqcup_f X$. Next, consider $Z=\text{pt}$ and the mapping $g:Y\to Z$. So, we have $$Z\sqcup_g \big(Y\sqcup_f X\big)\cong Z\sqcup_{g\circ f}X$$ by Law of Horizontal Composition. Next, note that both $g$ and $g\circ f$ is constants. Now, adjunction space of constant map is quotient space, so that $$\frac{Y\sqcup_f X}{Y}\cong Z\sqcup_g \big(Y\sqcup_f X\big)\text{ and }Z\sqcup_{g\circ f}X\cong\frac{X}{A}=\frac{\sqcup_\alpha\Bbb D^n}{\sqcup_\alpha\Bbb S^{n-1}}\cong\lor_\alpha\Bbb S^n$$$$\implies\frac{Y\bigsqcup_f \sqcup_\alpha\Bbb D^n}{Y}\cong\lor_\alpha\Bbb S^n.$$

  • $\begingroup$ I think you meant reduced homology for the wedge and quotient space. Anyway, thank you for the answer. $\endgroup$
    – Rundasice
    Jan 14, 2021 at 13:24
  • $\begingroup$ Yeah, I have written down everything up to homeomorphism, after that you can do it now. Edited. $\endgroup$
    – Someone
    Jan 14, 2021 at 13:31

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