# Finding coarser topology and homeomorphism between R and two tangent circles

Let $\tau$ be the usual topology on the real line $\mathbb{R}$. Does there exists a topology $\tau_{0} \subset \tau$ such that $(\mathbb{R},\tau_{0})$ is homeomorphic to the figure eight? Also, is it possible to find a topology $\tau_{0} \subset \tau$ and a quotient space with this topology such that this space is homeomorphic to $\mathbb{R}$?

What's the trick for this one?

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@Theo Buehler: We can identify the two end points of 0 and 1 and to get a circle. But $\mathbb{S}^{1}$ is not homeomorphic to $\mathbb{R}$. Can you please explain a little bit more your hint? –  student Jan 20 '11 at 21:57
Decide that $0 \in \mathbb{R}$ should be mapped to the point with a crossing. How should you define the neighborhoods of $0$? Moreover, what do you know about quotient spaces of compact spaces? –  t.b. Jan 20 '11 at 21:58
In other words: choose $0$ as the crossing point and parameterize the figure eight by the reals in such a way that the ends of the real line approach $0$ again. –  t.b. Jan 20 '11 at 21:59
@Theo Buehler: Thanks, I will try this. –  student Jan 20 '11 at 22:11
We let $\tau_{0}$ be the subset of open sets of $\tau$ consisting of $U$ in $\tau$ so that if $U$ contains $0$ then $U$ contains both some interval $(a,\infty)$ and some interval $(-\infty,b)$. This essentially glues $\infty$ and $-\infty$ to $0$ giving a figure eight.
$0$ is the intersection of the circles, $\mathbb{R}^{+}$ is one branch and $\mathbb{R}^{-}$ is the other.