TITLE: Homotopy equivalence of a space with the sphere
QUESTION [13 upvotes]: I have some trouble with the following problem.
A space $X$ is obtained by gluing two $2$-cells to a circle $S^1$ using maps winding $2$-times and $3$-times around $S^1$. Show that $X$ is homotopy equivalent to $S^2$.

REPLY [2 votes]: Denote $2$-cells of $X$ by $D_1$ and $D_2$. There are glueing maps $\partial D_1\to S^1,z\mapsto z^2$ and $\partial D_2\to S^1,z\mapsto z^3$.
Consider the cellular decomposition of $S^2$ with two $2$-cells $D_1'$ and $D_2'$.
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Let $f_1:{D_1'}\to D_1$ be $z\mapsto z^3$, and $f_2:{D_2'}\to D_2$ be $z\mapsto z^2$. As it easy to see, the restrictions of $f_1$ and $f_2$ to the equator coincide, so we have well-defined map $f:S^2\to X$. This map is homology equivalence and $X$ is $1$-connected, therefore $f$ is a homotopy equivalence.