# Is this quotient space a 1-manifold?

Let $R$ be the real line, $R'$ an isomorphic copy of the real line and $\phi : R \rightarrow R'$ an isomorphism. Consider the quotient space $X$ of $R \cup R'$ that results from the equivalence relation $x \sim x' \iff x' = \phi(x)$ and $x \neq 0$.

Is $X$ a 1-manifold? Or, is it a topological space in which every point has a neighborhood homeomorphic to $\Bbb{R}$?

I currently have no idea, even though I suspect there's a problem at $x=0$. Any hints and ideas will be appreciated.

• One can check that every point of $X$ has a neighborhood homeomorphic to $R$, but this is not the only condition required for $X$ to be a manifold. What are the other(s)? – Travis Willse Apr 11 '17 at 22:26
• Just take $\phi$ to be the identity and you get a classic counterexample, the line with two origins. – Alex Provost Apr 11 '17 at 22:33
• If we denote by $\pi$ the quotient map $R \amalg R' \to X$, by definition a set $U \subset X$ is open iff $\pi^{-1}(U)$ is open in $R \amalg R'$. In particular, if $V \subset R$ is open, then $\pi(V) \subset X$ is open, too. – Travis Willse Apr 12 '17 at 21:55
• You're welcome. To answer your question $\pi(V)$ is a subset of $X$, not of $R \amalg R'$. On the other hand, if $U \subset X$ does not include either of the distinguished points, then $\pi^{-1}(U)$ is the disjoin union of $W := \pi^{-1}(U) \cap R$ and its isomorphic image $\phi(W)$. – Travis Willse Apr 12 '17 at 23:02
• There isn't much to be gained, by the way, by letting $\phi$ be a general isomorphism (in whatever category). For topological purposes, there's no harm in just taking $\phi$ to be the identity. It might be easier, at least for purposes of actually describing explicitly images and preimages, to reframe things slightly and think of $R \amalg R'$ as $\Bbb R \times \{0, 1\}$ and declare $\sim$ to be the equivalence relation characterized by $(x, 0) \sim (x, 1)$ for $x \neq 0$. – Travis Willse Apr 12 '17 at 23:05

• Thanks, this helped. However, I am not sure what is the topology on $X$ (or, how to construct a homeomorphism from a neighborhood of a point to an open interval, because I'm not sure what a neighborhood of a point in this space looks like). If $\phi$ was the identity function, it would be clear (for example as in the answer here: math.stackexchange.com/questions/1038320/…) but the fact that $x$ is glued to $\phi(x)$ affects the definition of neighborhoods of this quotient space, doesn't it?... – Whyka Apr 12 '17 at 19:08
• @Whyka It actually doesn't matter at all. Since $\phi$ is an isomorphism (whatever you mean by that - homeomorphism? diffeomorphism?), what you get by gluing $x$ to $\phi(x)$ is totally equivalent to what you get by gluing $x$ to $x$. – Amitai Yuval Apr 12 '17 at 19:29
• What do you mean by "equivalent"? If I denote $\Bbb{R}_{1}\cup\Bbb{R}_{2}$ by $\Bbb{R}\times\{1,2\}$, can I still say (for example) that for a point $(a,i)$, $a\in\Bbb{R}-0$ and $i\in\{1,2\}$, a neighborhood excluding 0 is, $\{(x,i):x\in(a-\epsilon,a+\epsilon),\text{ }\epsilon=\frac{1}{2}|a|\}$? Where does the fact $(x,1)\sim(\phi(x),2)$ go? – Whyka Apr 12 '17 at 19:45
I assume $\phi$ is a linear isomorphism, thus $\phi(0)=0'$. $X$ is not a manifold since it is not separated. If in the definition of a manifold you consider only separated spaces: Let $p:R\bigcup R'\rightarrow X$ be the quotient map. You cannot find neighborhoods of $U$ of $p(0)$ and $U'$ of $p(0')$ such that $U\cap U'$ is empty. But every of its point has a neighborhood homeomorphic to $\mathbb{R}$.