# Computing homology groups of some quotient space (Hatcher 2.2.13)

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 $$$$0 \longrightarrow \mathbb{Z} \longrightarrow H_2(X/(S^1 \cup_2 e^2_1)) \longrightarrow \mathbb{Z}_2 \longrightarrow 0 \longrightarrow H_1(X/(S^1 \cup_2 e^2_1)) \longrightarrow 0.$$$$ Hence, we deduce that $$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)$$.

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 $$\frac{Y\sqcup_f X}{Y}\cong Z\sqcup_g \big(Y\sqcup_f X\big)\text{ and }Z\sqcup_{g\circ f}X\cong\frac{X}{A}=\frac{\sqcup_\alpha\Bbb D^n}{\sqcup_\alpha\Bbb S^{n-1}}\cong\lor_\alpha\Bbb S^n$$$$\implies\frac{Y\bigsqcup_f \sqcup_\alpha\Bbb D^n}{Y}\cong\lor_\alpha\Bbb S^n.$$