Is $[0,1]/{\approx}$, where $0\approx \frac{1}{2}$, homeomorphic to $S^1$ 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 $S^{1}$.)
 A: 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 $\left(0, \frac{1}{2}\right) \cup \left(\frac{1}{2}, 1\right]$.

A: $[0,1]/\approx$ 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.
A: $\hspace{10mm} \hspace{10mm} \hspace{10mm}$

A: 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 $\frac{1} {2} +  \epsilon$
A: 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.
