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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.

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    $\begingroup$ 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. $\endgroup$ Commented May 3, 2018 at 10:12

4 Answers 4

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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 $y$.

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  • $\begingroup$ 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 $s,t\in\{0,1\}$. $\endgroup$
    – Christoph
    Commented May 3, 2018 at 8:50
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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.

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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.

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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).

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