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Consider the quotient space ${[0,1]}/{\sim}$, where $[0,1]$ is equipped with the standard Euclidean topology, and where and the equivalence relation $\sim$ identifies its endpoints, i.e. gives rise to a partition $$ \left\{ \left\{ x\right\} :x\in(0,1)\right\} \cup\{0,1\} $$ of $[0,1]$.

Now it is clear that $[0,1]/{\sim}$ is homeomorphic to the circle $S^{1}$. My question is: Suppose we replace the equivalence relation with another, $\approx$, which identifies a different pair of points, say $0$ and $\frac{1}{2}$; it would thus now give rise to the partition $$ \{ \left\{ x\right\} :x\in(0,1]\setminus\{ \frac{1}{2}\} \} \cup\{0,\frac{1}{2}\}. $$

Is then ${[0,1]}/{\approx}$ still homeomorphic to $S^{1}$?

(Note: I suppose that's not the case, due to some high-level theorems that immediately tell us, that it can't be the case; but I don't know those theorem.
For whose interested, here's a sketch of the proof of the claim about $[0,1]/{\sim}$:

The map $$ f:[0,1]\rightarrow S^{1},\ t\mapsto(\cos2\pi t,\sin2\pi t) $$ is a continuous surjection that makes the same identifications as the canonical projection $p$ from $[0,1]$ to $[0,1]/{\sim}$, i.e. $p(x_{1})=p(x_{2})$ iff $f(x_{1})=f(x_{2})$. Some high-level theorem then guarantee us that $f$is a quotient map, so $[0,1]/{\sim}$ is homeomorphic to $S^{1}$.)

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    $\begingroup$ Your space looks like a baloon with a rope ;) You glued the end of a rope to its middle. It cannot be homeomorphic to $S^1$ because the image of $1/2$ haven't got euclidean neighborhood. $\endgroup$
    – Leonhardt von M
    Apr 9, 2015 at 17:06
  • $\begingroup$ Maybe a bit more intuitively: after removing point where the baloon meets the rope one obtains two components while after removing any point of $S^1$ only one component remains. $\endgroup$
    – Adam Baranowski
    Apr 9, 2015 at 20:00

5 Answers 5

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Hint Denote $X := [0, 1] /\!\approx$. If $\phi: X \to S^1$ is a homeomorphism, then so is $$\phi\vert_{X - \{\ast\}} \to S^1 - \{\phi(\ast)\}$$ where $\ast$ is the point in $X$ produced by identifying $0$ and $\frac{1}{2}$. Is this possible?

Additional hint Since the only nonsingleton equivalence class of $\approx$ is $\left\{0, \frac{1}{2}\right\}$, $X - \{\ast\}$ is homeomorphic to $\left(0, \frac{1}{2}\right) \cup \left(\frac{1}{2}, 1\right]$.

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$[0,1]/\approx$ is homeomorphic to a number $6$. It has one endpoint (a point whose removal does not disconnect any of its neighbourhoods), and one with a neighbourhood homeomorphic to a $T$ (the union of three line segments joined at a point, where the removal of that point makes the number of connected components go from $1$ to $3$). $S^1$ has neither of these.

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  • $\begingroup$ This is a very clear answer. Shouldn't, though, the number of connected components go from 1 to 2 ? $\endgroup$
    – user36675
    Apr 11, 2015 at 10:16
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    $\begingroup$ The removal of the point makes the number of connected components of the whole space go to 2, but makes the number of connected components of the neighbourhood go to 3. $\endgroup$
    – Robert Israel
    Apr 12, 2015 at 4:47
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If Nicolas's answer (which I upvoted) isn't enough, assume you had a homeomorphism and look at what would happen to $\epsilon$, $\frac{1} {2} - \epsilon$ and $\frac{1} {2} + \epsilon$

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We have a completely rigorous answer and a picture. Here's something in between.

View $[0,1]$ as $[0,1/2] \cup (1/2,1]$. When you take the quotient on these two pieces, you get $S^1$ and $(1/2,1]$ respectively. So the open neighborhoods of every point except $1/2$ are specified this way. For $1/2$, all open neighborhoods contain points both in the circle and in the remaining interval. So you indeed have a circle with an interval attached to it at one point, as Nicolas' picture shows.

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