# How to show that the space 'line with two origins' is path connected?

Consider the equivalence relation on $\mathbb R \times \{0,1\}$ that identifies $(x,0)$ with $(x, 1)$ whenever $x \neq 0$. Let $L$ be the quotient space. This space is called line with two origins.

From the picture of it, I understand that it is path connected but analytically how can we show it?

Any help is appreciated. Thank you.

• This space is also interesting since its fundamental group at any point is the integers. This can be proved, as for the circle, by using the van Kampen Theorem with a set of two base points. Commented May 3, 2018 at 10:12

This isn't complicated, but there's a lot of casework.

If $x,y<0$ or $x,y>0$, just take the line segment connecting them.

If $x<0<y$, just take the line segment, passing through your favorite zero.

If $x\neq0$, $y$ is one of the zeros, just take the line segment connecting $x$ to $0$.

If $x$ and $y$ are distinct zeros, start at $x$, go to $\frac{1}{2}$, turn around, and end at $y$.

• Actually no casework needed: A path from $[(x,t)]$ to $[(y,s)]$ would be two straight lines: $[(x,t)]$ to $[(1,t)]$ followed by $[(1,s)]$ to $[(y,s)]$, where we note that $(1,t)\sim(1,s)$ for all $s,t\in\{0,1\}$. Commented May 3, 2018 at 8:50

We can easily find a path between two points which is not origins. For example, if $a$ is negative and $b$ is positive then just consider the interval from $a$ to $b$ which passes one of two origins.

A path between an origin and non-origin also can be finded easily. The problem is the path between two origins, but it turns out be not so hard: Start from an origin and draw a path from that origin to 1. Then draw another path from 1 to the other origin. You can write down the function defining the path explicitly. Its continuity will follows from the fact that joining two continuous functions is also continuous.

Another approach would be to use the fact that $L$ is also the quotient of $X = (\mathbb{R} \times \{0,1\}) \cup (\{1\}\times[0,1])$ where the quotient maps the entire interval $\{1\} \times [0,1]$ to the same point (i.e., the same as the image of $(1,0)$ and $(1,1)$).

The space $X$ can much more easily be shown to be path connected and then you can use the fact that the continuous image of a path connected space is path connected.

Here we make a general argument.

Recall that any topological space $X$ can be partitioned into path components and that if the path component containing a point is the whole space, then $X$ is path connected.

Proposition 1: Let $X$ be a topological space and $f: \mathbb R \to X$ and $g: \mathbb R \to X$ two continuous functions satisfying the following condition:

$\tag 1 X = f(\mathbb R) \cup g(\mathbb R) \text{ and } f(\mathbb R) \cap g(\mathbb R) \ne \emptyset$

Then $X$ is path connected.

Proof

Let $x_0$ be any point in the intersection of the two ranges and consider another arbitrary point $x_1 \in X$ that is distinct from $x_0$. Now $x_1 \in f(\mathbb R)$ or $x_1 \in g(\mathbb R)$. If we assume that $x_1 \in f(\mathbb R)$, select real numbers $a_0$ and $a_1$ such that $f(a_0) = x_0$ and $f(a_1) = x_1$. Clearly if we restrict $f$ to the closed interval defined by $a_0$ and $a_1$ we get a path connecting $x_0$ to $x_1$.

An identical argument shows that a path can be created if $x_1 \in g(\mathbb R)$.

We have shown that we can path connect $x_0$ to any other point, so $X$ is indeed path connected.$\quad \blacksquare$

The OP can apply proposition 1 to their problem by setting up both $f$ and $g$ as the composition of two continuous functions (in a natural way).