Show that this gluing of spaces produces the circle.

Question

I'm trying to understand this example here:

Help understanding CW-complex construction.

So basically, we consider a point $x_0 \in \mathbb{R}$ and we consider the set

$$X^1 := \{x_0\} \sqcup [0,1]/\sim$$
where $\sim$ is the equivalence relation identifying $0 \sim 1 \sim x_0$.

It is claimed that $X^1$ is homeomorphic to the circle $S^1$, but I can't see why this is true.

So, let me be really formal about my understanding. We have a disjoint union, so we will give labels.

$$\{x_0\} \sqcup [0,1] := \{(x_0,1)\} \cup \{(y,2): y \in [0,1]\}$$

and then I guess we identify $(x_0,1) \sim (0,2) \sim (1,2)$. How does this give something homeomorphic to $S^1$?

I guess I will need to view $S^1$ as the quotient space $\mathbb{R}/\sim$ where $x \sim y \iff x-y \in \mathbb{Z}$ and pick a representative $x \in [0,1[$ of $S^1$ and define

$$\psi: S^1 \to X^1: [x] \mapsto \{(x,2)\}$$

I can see this is a bijection. How can I show that this is continuous? Since $S^1$ is compact, it will follow that $\psi$ is a homeomorphism.

Answer 1

I would suggest that you be "really formal" in a somewhat different sense: let $X^1 = \{-1\} \cup [0,1]$ with equivalence relation $-1 \sim 0 \sim 1$.

Second, I would suggest that you define a map which goes straight from $X^1$ to $S^1 \subset \mathbb R^2$, without passing through the intermediate step of viewing $S^1$ as the quotient $\mathbb R / \sim$. So, just define $$f : X^1 \to S^1 $$ by $f(t)=(\cos(2\pi t), \sin(2 \pi t))$.

Now use a simple theorem from topology:

If $f : Y \to Z$ is a continuous, surjective function from a compact space $Y$ to a Hausdorff space $Z$, and if $\sim$ is the equivalence relation on $Y$ defined by $y_1 \sim y_2$ if and only if $f(y_1)=f(y_2)$, and if $Y / \sim$ is the quotient space of this equivalence relation, and if $F : Y /\!\sim \, \to Z$ is the induced function defined by $F[y]=f(y)$, then $F$ is a homeomorphism.