$ widehat {(0,1)} = (0,1) cup {p } $, or $ p not in (0,1) $. A set $ U $ in $ widehat {(0,1)} $ is open if ($ U not nor p $ and $ U $ is open in $ (0.1) $) or ($ U ni p $ and $ (0,1) setminus U $ is a compact subset of $ (0.1) $).

CA watch $ widehat {(0,1)} $ is homeomorphic to $[0,1]/ sim $ or $ x sim y ssi x = y text {or} {x, y } = {0,1 } $.

A set of representatives of $[0,1]/ sim = {[x] mid x in[0,1] } $ is $ bigcup_ {x in (0,1)} underbrace {[x]} _ {= {x }} cup underbrace {[0]} _ {=[1]= {0,1 }} $.

Let $ p:[0,1] twoheadrightarrow [0,1]/ sim: x mapsto [x]$ to be the quotient map, so $ U subset[0,1]/ sim $ is open if $ p ^ {- 1} (U) subset [0,1]$ is open.

I have to prove that

$$ begin {align *} f colon[0,1]/ sim & longrightarrow widehat {(0,1)} \ [x]& longmapsto x text {if} x in (0,1) \ [0]=[1]& longmapsto p end {align *} $$

is a homeomorphism.

It is clear that it is a bijection. I have problems to prove that it is continuous.

Let $ U subset widehat {(0,1)} $ open. assume $ p not in U $then $ U $ is open in $ (0.1) $, So $ U = (0,1) cap B $ ($ B $ open in $ mathbf {R} $). then $ f ^ {- 1} (U) $ is open in $[0,1]/ sim $, since $ p ^ {- 1} (f ^ {- 1} (U)) = U $which is open in $[0,1]$, since $ U = underbrace {((0,1) cap B)} _ { text {open in} mathbf {R}} cap [0,1]$.

Now assume $ to U $. then $ (0,1) setminus U $ is a compact subset of $ (0.1) $ (I guess that means by Heine Borel that he's closed $ (0.1) $?) So $ (0,1) setminus U = (0,1) cap C $ ($ C $ shut in $ mathbf {R} $). Now $ (0,1) setminus ((0,1) setminus U) = (0,1) cap underbrace {( mathbf {R} setminus C)} _ { text {open in} mathbf {R}} $. It means that $ p ^ {- 1} (f ^ {- 1} (U)) $ is the union of an open subset of $ (0.1) $ and $ {0,1 } $, which is not open in $[0,1]$.

I think this problem is difficult for me because I do not feel familiar with the topology of the quotient yet. Could someone provide some help or a simpler way to handle this?