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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?

user198182
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1 Answers1

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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 $S^2 \underset{\text{deg}\, =\,2}{\xrightarrow{\phantom{\text{deg}\, =\,2}}} S^2$.

user149792
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    Could you help with constructing the homotopy between the composition $S^2 \rightarrow X \rightarrow S^2$ and $id_{S^2}$? I am not sure how to check this. – user198182 Mar 13 '15 at 16:40