Problem 1.1 (Line with two origins)

Let $\pi :X\to M$ be the natural projection sending each point to its equivalence class. This map defines the quotient topology of $M$. Let $p_1$ and $p_2$ denote the upper origin $[(0,1)]$ and the lower origin $[(0,-1)]$ of $M$ respectively.

$\pi$is an open map. Recall the following basic result about quotient maps.

Lemma. A quotient map $\pi$is an open map if and only if

This property holds in the case of the line with two origins. To see why, fix an open subset $U\subset X$. Then it is easy to see that ${\pi }^{-1}(\pi (U))=U\cup U^R$ where $U^R$ denotes the reflection of $U$ about the $x$-axis. Both sets are open in $X$, making $\pi$ an open map.

$M$is second countable. Define an “open interval” in $M$ to be the image of the corresponding open interval in $ℝ$ under the quotient map $\pi$:

These “open intervals” are open in $M$ since $\pi$ is an open map. Imitating the basis $\{(-q,q)\mid q\in {\mathbb{Q}}_{\text{+}}\}$ of the real line, we can define a countable basis $B$ of $M$ as:

It is easy to verify that $B$ indeed constitutes a basis for $M$.

$M$is locally Euclidean. For each point $[p]\in M$, we should find an open neighbourhood $U\subset M$ containing $[p]$ for which there is a homeomorphism $\phi :U\to ℝ$ (cf. Exercise 1.1).

The trick is to notice that the upper part of $X$ can generate the quotient line with the upper origin $p_1$ and the lower part of $X$ can generate the quotient line with the lower origin $p_2$ (i.e., any point $[p]\in M$ is contained in $\pi (ℝ\times \{1\})=M\setminus \{p_2\}$ or in $\pi (ℝ\times \{-1\})=M\setminus \{p_1\}$). In other words, two large sets already satisfy the desired open neighbourhood condition: $M\setminus \{p_2\}$ and $M\setminus \{p_1\}$ . Both sets are open since $\pi$ is open.

Such separation results in additional properties. By construction, we can find at least one $x$ such that $p\sim (x,1)$ for all $p\in M\setminus \{p_2\}$ and $p\sim (x,-1)$ for all $p\in M\setminus \{p_1\}$. Conversely, if there is one more $x^{\prime }$ with $p\sim (x^{\prime },1)$ or $p\sim (x^{\prime },-1)$ respectively, we must necessarily have $x=x^{\prime }$. This allows us to write our equivalence classes $[p]$ in either $M\setminus \{p_2\}$ or $M\setminus \{p_1\}$ as $[x]$ for some scalars $x\in ℝ$ without any problems (which is the first hint that our manifold is one-dimensional). Using this property, we define:

and

These functions are

• (bijective) ${\phi }_1$ is obviously surjective. To show injectivity, suppose that ${\phi }_1([x])={\phi }_1([x^{\prime }])$ for some $[x],[x^{\prime }]\in M\setminus \{p_2\}$. Then $x=x^{\prime }$ and so $[(x,1)]=[(x^{\prime },1)]$. A similar argument works for ${\phi }_2$.
• (continuous) Let $\tilde{U}$ be open in $ℝ$. The ${\varphi }_1^{-1}(U)$ maps to $\pi (U\times \{1\})$ and ${\varphi }_2^{-1}(U)$ to $\pi (U\times \{-1\})$ - both open since $\pi$ is an open map.

$M$is not Hausdorff. We argue that the points $[(0,1)]$ and $[(0,-1)]$ are not separable in $M$. Suppose for the sake of contradiction that we can find two open sets $U$ and $U^{\prime }$ containing $[(0,1)]$ and $[(0,-1)]$ respectively. Since $\pi$ is continuous, the sets ${\pi }^{-1}(U)$ and ${\pi }^{-1}(U^{\prime })$ are two open subsets of $X$ containing $(0,1)$ and $(0,-1)$ respectively. We can then find $𝜀>0$ small enough so that

In other words, since the image of the preimage is a subset of the original set,

- a contradiction to the fact that $U\cap U^{\prime }=\varnothing$.