Is this Space Homotopy Equivalent to $S^2$

Question

Let $X$ be the space $S^1$ with two $2$-cells attached via maps of relatively prime degrees. This space is simply connected and has the homology of $S^2$, but is it homotopy equivalent to $S^2$?

Answer 1

By Hurewicz theorem, $\pi_2(X) \cong \Bbb Z$. Let $f : S^2 \to X$ be a generator of $\pi_2(X)$. This map is of degree one by the same theorem and hence induces isomorphisms on all homology groups. Since $X$ is a CW complex and both $X$ and $S^2$ are simply connected, it follows that $f$ is a homotopy equivalence by the homology version of Whitehead's theorem. For a statement and proof of this theorem, see corollary 4.33 in Hatcher's Algebraic Topology.

Answer 2

Take one of the 2-cells in $X$ and call it $A$. Then $(X,A)$ is a CW-pair, hence it has the homotopy extension property. Since $A$ is contractible, it then follows that the quotient map $\pi: X \rightarrow X/A$ is a homotopy equivalence. But $X/A$ is homeomorphic to $S^2$.

Answer 3

There is a solution here http://www.math.ku.dk/~moller/blok1_05/AT-ex.pdf?q=allen-hatcher page 25

Briefly, it can be stated like this: $Y^0={e_0}$, $Y^1=S^1$, $Y^2=Y$, where we denote the $1$-cell: $e_1$ and the $2$-cells $e_a$ and $e_b$. where $a,b$ are relatively primes $\pi_1(S^1\cup e_a, e_0)=\{e_1|(e_1)^a\}$. The attaching map $f_a:S^1\rightarrow S^1\subset S^1\cup e_a$ is an element in $\pi_1(S^1\cup e_a, e_0)$ and $[f_a]=e_1^a=e_1$. Thus, the attachment map is homotopic to the degree one attachment map $id:S^1\rightarrow S^1\subset S^1\cup_{id} e_a$. Let $0:S^1\rightarrow S^1$, $0(S^1)=e^0$ be the constant map. We have the following homotopies: $Y=S^1\cup_{f_a}e_a\cup_{f_b}e_b\sim S^1\cup_{f_a}e_a\cup_{id}e_b=S^1\cup_{id}e_b\cup_{f_a}e_a=D_2\cup_{f_a}e_a \sim D^2\cup_0 e_a=D^2\vee S^2\sim S^2.$ The homotopies rely on 0.18 from Hatcher's book.