I am working on the following problem:
Let $X$ be the CW complex obtained from $S^1$ by attaching two $2$-cells: one by a map of degree $2$, and one by a map of degree $3$. (a) Compute the homology groups $H_k(X;\mathbb{Z})$ for all $k \geq 0$. (b) Show that $X$ is homotopy equivalent to $S^2$. (c) What conditions on the integers $m$ and $n$ ensure that the answers to parts (a) and (b) remain the same if the attaching maps have degrees $m$ and $n$?
Here are my solutions to parts (a) and (b):
(a) Since $X$ is a $2$-dimensional CW complex (with one $0$-cell $p$, a $1$-cell $\gamma$, and two $2$-cells $\alpha$ and $\beta$ attached via maps of degrees $2$ and $3$, respectively), $H_n(X)$ is trivial for all $n \geq 3$. Further, $H_0(X) \cong \mathbb{Z}$ by path-connectedness of $X$. We use cellular homology to compute the remaining homology groups $H_1(X) \cong H_1^{\text{CW}}(X)$ and $H_2(X) \cong H_2^{\text{CW}}(X)$. We have the cellular chain complex $0 \xrightarrow{d_3} C_2(X) = \langle \alpha,\beta \rangle \xrightarrow[]{d_2} C_1(X) = \langle \gamma \rangle \xrightarrow[]{d_1} C_0(X) = \langle p \rangle \xrightarrow[]{d_0} 0$. By the cellular boundary formula, $d_2(\alpha) = \pm 2\gamma$ and $d_2(\beta) = \pm 3\gamma$, while $d_1$ is the zero map. Therefore, $H_1(X)$ is trivial and $H_2(X) \cong \mathbb{Z}$.
(b) Let $C(m)$ denote the $2$-complex obtained from $S^1$ with its usual cell structure by attaching a $2$-cell by a map of degree $m$, and let $C(m,n) = C(n,m) = C(n)(m) = C(m)(n)$ denote the $2$-complex obtained from $S^1$ with its usual cell structure by attaching two $2$-cells by maps of degrees $m$ and $n$, respectively. Note that $S^2 = C(1,1)$, $X = C(2,3)$, $C(2) \simeq \mathbb{R}P^2$, and $C(1) = D^2$. Further, recall that two CW complexes are homotopy equivalent when they have the same cell structure and each of the corresponding attaching maps are homotopic. Since $D^2$ is contractible, $\pi_1(D^2)$ is trivial. Therefore, $C(2,1) = C(1)(2) \simeq C(1,1) = C(1)(1)$. Meanwhile, since $\pi_1(\mathbb{R}P^2) \cong \mathbb{Z}_2$, $C(2,1) = C(2)(1) \simeq C(2,3) = C(2)(3)$. Therefore, we obtain $C(2,3) \simeq C(1,1)$, as desired.
Now, I'm struggling with part (c). The answer is supposed to be for gcd(m,n) = 1, but I can't seem to reach this conclusion.
I believe that, for any $m$ and $n$, the computations in (a) will remain the same -- the only homology group that could potentially differ for general $m$ and $n$ is $H_2(X)$, but I argue that $H_2(X) \cong \mathbb{Z}$ for any $m$ and $n$, since there will still be some linear combination of the two $2$-cells that maps to $0$ by $d_2$. Also, for any $m$, we have the homotopy equivalence $C(1,1) \simeq C(m,1)$ (again due to the fundamental group of $C(1) = D^2$ being trivial). So, the answer to (c) seems to boil down to what condition on $n$ yields the homotopy equivalence $C(m,1) \simeq C(m,n)$. Again, the condition is supposed to be that $n$ is coprime to $m$, but I can't quite see this. How to see why this condition is necessary?
We want the homotopy equivalence between $C(m)(1)$ and $C(m)(n)$. We do know that $C(m)$ is exactly the Moore space $M(\mathbb{Z}_m,1)$ -- so we want a homotopy equivalence between the space obtained from this Moore space by attaching a $2$-cell via a map of degree $1$ and the space obtained from this Moore space by attaching a $2$-cell via a map of degree $n$. Why would this homotopy equivalence hold only when $n$ is coprime to $m$?
Thanks!