Is the union of two circles homeomorphic to an interval? Let $Y$ be the subspace of $\Bbb R^2$ given by $Y=\{(x,y): x^2+ y^2=1\}\cup \{(x,y): (x−2)^2+ y^2=1\}$. Is $Y$ homeomorphic to an interval?
I have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $Y$. Basically if we remove a point from $Y$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?
 A: A cutpoint of a connected space $X$ is a $p \in X$ such that $X\setminus\{p\}$ is disconnected.
If $f:X \to Y$ is a homeomorphism of connected spaces $X$ and $Y$ and $p$ is a cut point of $X$ then $f(p)$ is a cutpoint of $Y$ (and vice versa).
If we take $X$ to be an interval, then $X$ has at most two non-cutpoints. (the endpoints in the case of a closed interval). $Y$ on the other hand has infinitely many non-cutpoints (all points except $(1,0)$). So there can be no homeomorphism between them by the observations in the second paragraph.
A: Here's an alternative argument that covers wide range of spaces, regardless of cut points.
Let $X$, $Y$ be any topological spaces. Consider a homeomorphism $f:X\to Y$. Now let $Z$ be any topological space and $\alpha:Z\to X$ be a continuous function. Then $\alpha$ is injective if and only if $f\circ\alpha$ is injective. Which is easy to see by applying $f$ to $\alpha$ and $f^{-1}$ to $f\circ\alpha$.
In particular $X$ admits an injective map $Z\to X$ if and only if $Y$ admits an injective map $Z\to Y$.
This shows that your $Y$ (or more generally any space containing $S^1$ as a subspace) cannot be homeomorphic to the interval $[0,1]$. Because there is an obvious injective continuous map $S^1\to Y$ while there is no continuous injective map $S^1\to [0,1]$. That's because every map $S^1\to [0,1]$ arises from a map $[0,1]\to [0,1]$ with the same values at endpoints and so by the intermediate value property it cannot be injective on $(0,1)$, which then "lifts" to $S^1$.
A: Just for a change: Let $D$ be the open disk in $\Bbb R^2$ centered at $p=(1,0)$ with radius $1$ and let $E=D\cap Y.$ Then $E$ is a connected subspace of $Y$  and $E\setminus \{p\}$ is the union of $4$ pairwise-disjoint non-empty connected open subspaces of $E$. But if $E'$ is any connected subspace of $\Bbb R$ and $p'\in E'$ then $E'\setminus \{p'\}$ is either connected or is the union of $2$ disjoint non-empty connected open subspaces of $E'$.
