Proving $[0,2]\big/[1,2]$ is homeomorphic to $[0,1]$

Question

Prove $[0,2]\big/[1,2]$ is homeomorphic to $[0,1]$.

Where $A\big/B$ is the quotient set where all points of $B$ are identified as a single point (and other remain distinct). Our definition of a homeomorphism $i:X\to Y$ amounts to a bijective map between topological spaces such that the topology of $X$ is exactly

$$\left\{ i^{-1}(U):U\text{ open in }Y \right\}$$

and we have no other theorems to hasten the process. My problem is mostly with formality, where a lot of structure is apparently "lost" when moving to talk about representatives instead, and the discussion becomes problematic. My proof follows, but it feels like I'm doing something wrong for a relatively simple quotient:

Define $f:\left[0,2\right]\big/\left[1,2\right]\to\left[0,1\right]$ by

$$ f\left(\left[x\right]\right)=\begin{cases} x & x\in\left[0,1\right)\\ 1 & x\in\left[1,2\right] \end{cases}$$

to see this is well defined, if $\left[x\right]=\left[y\right]$ there are two cases, either $x,y\in\left[0,1\right)$ and then $x=y$, or $x,y\in\left[1,2\right]$ and are both mapped to $1$. Then f is injective, as $\left[x\right]\neq\left[y\right]$ means $x\neq y$ hence if $\left[x\right],\left[y\right]\neq\left[1\right]$, $$f\left(\left[x\right]\right)=x\neq y=f\left(\left[y\right]\right)$$ and if without loss of generality $\left[x\right]=\left[1\right]$, then $\left[y\right]\neq\left[1\right]$ and

$$f\left(\left[y\right]\right)=y\neq1\neq f\left(\left[x\right]\right).$$

It is surjective as for any $x\in\left[0,1\right], f\left(\left[x\right]\right)=x$.

Suppose $q:\left[0,2\right]\to\left[0,2\right]\big/\left[1,2\right]$ is the quotient map, then given an open interval (subbasis element) $\left(a,b\right)\subset\left[0,1\right]$, if $b\leq1$ then $$q^{-1}\left(f^{-1}\left(\left(a,b\right)\right)\right)=\left(a,b\right)$$

and if $b>1$,

$$q^{-1}\left(f^{-1}\left(\left(a,b\right)\right)\right)=\left(a,2\right] $$

showing that $f\circ q:\left[0,2\right]\to\left[0,1\right]$ is continuous, and therefore so is f. Similarly if $U\subset\left[0,2\right]\big/\left[1,2\right]$ is open, if $\left[1\right]\notin U$ then

$$f\left(U\right)=q^{-1}\left(U\right)=\left\{ x:\left[x\right]\in U\right\}$$

thus from the continuity of $q$, $f\left(U\right)$ is open, and if $\left[1\right]\in U$,

$$f\left(U\right)=q^{-1}\left(U\setminus\left[1\right]\right)\cup\left\{ 1\right\}$$

where $q^{-1}\left(U\setminus\left[1\right]\right)$ is open, as given any point $x$ it satisfies $x<1$, and given an open neighborhood $U_{x}$ of $\left[x\right]$ in U, if it contains $\left[1\right]$, then

$$q^{-1}\left(U_{x}\setminus\left[1\right]\right)=q^{-1}\left(U_{x}\right)\setminus\left[1,2\right]$$

which is open, showing $U\setminus\left[1\right]$ also contains such a neighborhood. We therefore only need to find a neighborhood for $1\in\left[0,1\right]$, but since $U$ contains a neighborhood for $\left[1\right]$, $q^{-1}\left(U\right)$ contains a neighborhood for $\left[1,2\right]$, which must contain an interval of the form $\left(a,2\right]$ for $a<1$, hence $\left(a,1\right]\in f\left(U\right)$ showing that $f\left(U\right)$ is indeed open and $f$ is a homeomorphism.

Answer 1

Here is a simplier argument using a classical lemma.

You have shown that $f$ is a continuous bijection. Then, we now that $[0,2]/[1,2]$ is compact, as it is a quotient of a compact space. Also, $[0,1]$ is Hausdorff, so any continuous bijection $[0,2]/[1,2]\to[0,1]$ is a homeomorphism.

You can find here a proof of the lemma (it is easy).

Answer 2

$f\left(\left[y\right]\right)=y\neq1\neq f\left(\left[x\right]\right)$

I guess you meant $f\left(\left[y\right]\right)=y\neq1=f\left(\left[x\right]\right)$.

if $b\leq1$ then $\ldots$ and if $b>1$ $\ldots$

It is $b<1$ and $b\geq1$ respectively.

$f\circ q$ is continuous, and therefore so is $f$

I don't remember this fact, but that is probably my memory's fault.

hence $(a,1]\in f(U)$

You mean $(a,1]\subseteq f(U)$.

With the corrections, your proof is correct from my point of view. And the "errors" you made could just be typos for all I know.

Answer 3

It's clear that $g: [0,2] \to [0,1]$ defined by

$$g= \begin{cases} x & \text{ if } x \le 1\\ 1 & \text{ if } x \ge 1\\ \end{cases} $$

is continuous by an application of the pasting lemma for the two closed sets $[0,1]$ and $[1,2]$ and $g$ restricted to the first is the identity and to the second the constant map $1$. So both are continuous on a closed part, they agree on the intersection $\{1\}$ and so the total function is continuous.

Your $f: [0,2]/{[1,2]} \to [0,1]$ is indeed well-defined and it obeys the property that $f \circ q=g$ (both sides are maps from $[0,2]$ to $[0,1]$) and this already shows $f$ is continuous. ($O$ open in $[0,1]$ implies that $g^{-1}[O]$ is open in $[0,2]$ and this equals $q^{-1}[f^{-1}[O]]$ and as $q^{-1}[f^{-1}[O]]$ is open, so is $f^{-1}[O]$ by the definition of the quotient topology).

So $f$ is continuous and bijective. The openness of $f$ needs a bit more work along the lines of what you did, distinguishing the case where $[1]$ is in the open set or not. There might be a smoother argument, perhaps.