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The circle $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}/\mathbb{Z}:=\{x+\mathbb{Z}\ |\ x\in [0; 1)\}$, but this answer states the topology is not the subspace topology. I wonder if it's the topology generated by half-open intervals $[a, b)$?

Also, what does the topology of $\mathbb{S}^1$ "look like", without embedding $\mathbb{S}^1$ in $\mathbb{R}^2$ or $\mathbb{C}$, but intrinsically? Is it made up of open intervals? Can we even think of open intervals in $\mathbb{S}^1$ without resorting to an "ambient space"?

Martin Sleziak
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étale-cohomology
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    Are you familiar with the quotient topology? – Amitai Yuval May 18 '17 at 23:31
  • Isn't f: [0,1) -> S^1, defined by f(x) = (cos(x), sin(x)) a bicontinuous function? If there exists a homeomorphism between the two spaces then they we say that they are homeomorphic. – Tucker May 18 '17 at 23:31
  • The quotient topology. The equivalence relation here is that $0$ and $1$ are in the same equivalence class, hence are identified. – avs May 18 '17 at 23:31
  • The subspace topology for what: $\mathbf S^1$, as a subset of $\mathbf R^2$? – Bernard May 18 '17 at 23:33
  • The answer is a bit misleading. Any space with the same cardinality as another space is homeomorphic to the other space if you just choose a different topology. In general when one speaks of $[0,1)$ it comes with the baggage of being a subspace of $\mathbb R$ with the usual topology. This gives $[0,1)$ its own "usual topology." In that topology it is not homeomorphic to the circle. $\mathbb R\mathbb Z$ also comes with a usual topology, the quotient topology, and in that topology it is homeomorphic to the circle. It so happens that the subset $[0,1)$ of $\mathbb R$ is mapped bijectively onto... – Matt Samuel May 18 '17 at 23:42
  • the circle by the quotient map, but not homeomorphically. – Matt Samuel May 18 '17 at 23:42

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First, it is important to note that $S^1$ is not a subspace on $\mathbb{R}$, and so, the subspace topology is irrelevant here. Rather, it is a quotient of $\mathbb{R}$.

As to what the topology looks like: locally, $S^1$ looks exactly like $\mathbb{R}$. This means that small intervals in the circle are homeomorphic to real intervals. The difference between the line and the circle is a global one. That is, it cannot be seen by small creatures living on a circle. (This is somewhat similar to the difference between $S^2$ and the plane. Recall that it took people many years before they realized we were all living on a sphere).

Edit: In the answer mentioned in the post, the circle is identified with the interval $[0,1)$. While this identification may help to understand $S^1$ as a set or as a group (in the same way one can identify the ring $\mathbb{Z}/m\mathbb{Z}$ with the set $\{0,\ldots,m-1\}$), I find it very confusing in the topological aspect. If I had to describe the topology of the circle by means of subsets of $\mathbb{R}$, I would take the closed interval $[0,1]$ and glue both its endpoints to one another.

Amitai Yuval
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  • Note that the quotient topology might affect local things to, like how you make a sphere out of a disc by identifying the entire edge to a single point. – Arthur May 19 '17 at 00:04
  • I think that by "subspace topology" they meant the subspace topology on $\mathbb{R}/\mathbb{Z}$ by identifying it with $[0,1)$. – Eric Wofsey May 19 '17 at 03:07
  • @EricWofsey Thanks, your comment made the post more clear to me. I edited my answer accordingly. – Amitai Yuval May 19 '17 at 04:56