# Double mapping cone of coprime maps between circles.

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:

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).

• I have no idea how to title this post. Please suggest/edit a helpful title. Jul 20 at 21:10
• 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. Jul 20 at 21:12
• @SammyBlack I'm considering $p$ and $q$ primes. Also, I totally agree that I can be horribly wrong. Jul 20 at 21:14

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 $$H_2 \text{Cone}(p, q) = \Bbb Z$$.