TITLE: Hatcher Exercise 2.2.13
QUESTION [5 upvotes]: In this exercise we have $X$, a 2-complex obtained from $S^1$ by attaching two 2-cells by maps of degree 2 and 3, respectively. I am trying to see why $X \simeq S^2$, in part b) of the exercise. I can't seem to construct a homotopy equivalence map. Maybe there is a way to show it without constructing the actual maps? Could you help?

REPLY [2 votes]: These two maps $S^2 \to X \to S^2$ give the homotopy equivalence $S^2 \simeq X$, as checked laboriously or using Corollary 4.33 in Hatcher.
If $A = \rho \cup \gamma \cup \beta$, then
$$
\require{AMScd}
\begin{CD}
X @>>> X/A \\
@A\cong AA     @V \cong VV  \\
S^2 @>>\text{deg}\, =\, 3> S^2
\end{CD}
$$and if $A = \rho \cup \gamma \cup \alpha$, then $X \to X/A$ is equivalent to $S^2 \underset{\text{deg}\, =\,2}{\xrightarrow{\phantom{\text{deg}\, =\,2}}} S^2$.