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Define an action of $\mathbb{Z}_2$ on $S^1$ by $(0,z)\mapsto z$ and $(1,z)\mapsto \bar{z}$. An orbit of $z$ is then the set $\{z,\bar{z}\}$. I claim the orbit space $S^1/\mathbb{Z}_2$ is homemorphic to $[-1,1]$.

Proof: Define $f:S^1\to [-1,1]$ by $f(z)=\Re(z)$. This is evidently continuous and if $z\sim z'$ then, for instance if $z'=\bar{z}$, we have $f(z)=f(\bar{z})$. Thus by the universal property of quotient maps, there exists a continuous function $\phi: S^1/\mathbb{Z}_2\to [-1,1]$ such that $\phi \circ \pi=f$ where $\pi$ is the canonical projection of $S^1\to S^1/\mathbb{Z}_2$. In fact, we have the luxury of writing $\phi([z])=\frac12(z+\bar{z})$. By a well known theorem, it is clear that $\phi$ is a homeomorphism, being a continuous bijection from a compact space $S^1/\mathbb{Z}_2$ (being the continuous image, $\pi(S_1)$), and the Hausdorff space $[-1,1]$.

But if instead you use an action that takes points to diametrically opposite points, i.e. $(0,z)\mapsto z$ and $(1,z)\mapsto -z$, then the corresponding orbit space is readily verified to be homeomorphic to $S^1$.

My three questions are 1) is the above proof and afterthought correct? and 2) is there a way to characterize all the orbit spaces of $S^1/G$ by various actions of $G=\mathbb{Z}_2$? and 3) Where would one want to begin studying algebraic topology if you come from an analysis background with knowing group theory, linear algebra, and basic point set topology? (like the most vital prerequisites, or fundamental concepts, is it appropriate to study Homology before homotopy, vice-versa, or concurrently? etc...). Sorry if the last two questions are too broad. Thanks!

Nap D. Lover
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Yes, your proof and afterthought are correct. I'm not sure if there is a way to characterize all the orbit spaces $S^1/\mathbb{Z}_2$ for various actions. Obviously, the resulting space will be compact and connected. Moreover, as a quotient of a locally compact Hausdorff space by a proper action, it will also be Hausdorff. Finally, if $\mathbb{Z}_2$ acts freely, as in your afterthought, then the quotient will be a compact $1$-manifold, i.e., another circle, and the quotient map will be a double covering map.

Alex Provost
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  • Thank you! Very informative. And The same argument can surely be generalized to prove $\mathbb{T}^k/\mathbb{Z}_2 \cong [-1,1]^k$, no? – Nap D. Lover Jul 23 '16 at 16:05
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    @LoveTooNap29 Yes, it is a straightforward generalization. It can even be visualized in the cases $k=1,2,3$. Just imagine $T^k$ as a quotient of the unit cube $I^k$ by the boundary relations. Then passage to the orbit space may be seen as folding $I^k$ across each coordinate hyperplane ${x_i = 0}$, $i=1,\ldots,k$. This process glues together the boundaries of $I^k$ which were to be identified on the torus, so the result is just $I^k$. – Alex Provost Jul 23 '16 at 17:22
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    By the way, if this subject interests you, Massey's "Algebraic Topology: An Introduction" is a very good starting point. Bredon's "Topology and Geometry" is fantastic and a good reference, but a bit dry. Finally, since you come from an analysis background, I'd be sure to check out Bott & Tu's classic "Differential Forms in Algebraic Topology." There the machinery of de Rham cohomology on differentiable manifolds is developed, and this can lead you to study e.g. Hodge theory to do analysis and differential equations on manifolds. Hope this helps! – Alex Provost Jul 23 '16 at 17:32