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Suppose we have two maps $f,g: S^1 \to S^1$ with $f(z) = z^p$, and $g(z) = z^q$ for two primes $p \neq q$. Consider the space $M$ formed by gluing together two copies of $D^2$ to a single $S^1$ along its boundary, according to the maps $f,g$. Here's a crude picture of what I mean: Sorry I only have access to MS paint right now

I want to show that $M$ is homotopy equivalent to the sphere $S^2$. How can I go about doing it?

Intuitively I can sort of see what's happening. On the left side, we're identifying points along a $p$-gon around the border of the disk, and on the right, we're doing the same for a $q$-gon. When we're gluing them together, since $(p,q) = 1$, I can see how everything gets identified to itself, and results in a $pq$-times folded circle $S^1$. But this is not at all a formal solution, and I don't even know if it's correct (because it's not a formal solution).

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  • $\begingroup$ I have no idea how to title this post. Please suggest/edit a helpful title. $\endgroup$ Jul 20 at 21:10
  • $\begingroup$ Note: you need a more delicate argument since with gluing maps of degrees $(p, q) = (2, 1)$, you get the projective plane $\mathbb{R}P^2$. It's not clear how you are using the fact that $p$ and $q$ are coprime. $\endgroup$ Jul 20 at 21:12
  • $\begingroup$ @SammyBlack I'm considering $p$ and $q$ primes. Also, I totally agree that I can be horribly wrong. $\endgroup$ Jul 20 at 21:14

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van Kampen implies that $\pi_1 \text{Cone}(f,g) = \Bbb Z/(p, q)$, where $(p,q)$ denotes the ideal generated by these (equivalently, the ideal generated by their greatest common denominator). In your case, this is the trivial group. I disagree with the commenter that the result is $\Bbb{RP}^2$ for $q = 1$ (in fact, in that case the result is homotopy equivalent to $S^2$ simply by collapsing the right-hand disc).

Now the cellular chain complex is $$\Bbb Z^2 \xrightarrow{\begin{pmatrix} p & q \end{pmatrix}} \Bbb Z \xrightarrow{0} \Bbb Z.$$ So long as one of $p, q$ is nonzero, the kernel of the first map is abstractly isomorphic to $\Bbb Z$, so $H_2 \text{Cone}(p, q) = \Bbb Z$.

The desired claim follows immediately from Hatcher 4C.1.

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  • $\begingroup$ Welcome to MSE. Are we all suppose to know what “Hatcher 4C.1” is? $\endgroup$ Aug 1 at 15:21

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