# Thread: Show quotient topology on [0,1] = usual topology on [0,1]

1. ## Show quotient topology on [0,1] = usual topology on [0,1]

Let X = [-1,1], and give X the usual topology.

Definte a function f:X -> [0,1] by f(x) = x^2. Show that the quotient topology on [0,1] induced by f is the same as the usual topology on [0,1].

This is so tricky. I figured out that I kinda have to prove that both are subsets of each other. So I need to say that the open subsets of f^-1(u) in [-1,1] lead to an open set in the usual sense of [0,1]. Then I need to say that the open subsets in [0,1] have open pre-images of f in X.

Trouble is, it's impossible to generalize an open subset. I think it's something like two half-open intervals at the end of the subset, with a union of open intervals in the middle?

Any help will be GREATLY appreciated.

2. Originally Posted by Throughpoint
Let X = [-1,1], and give X the usual topology.

Definte a function f:X -> [0,1] by f(x) = x^2. Show that the quotient topology on [0,1] induced by f is the same as the usual topology on [0,1].

This is so tricky. I figured out that I kinda have to prove that both are subsets of each other. So I need to say that the open subsets of f^-1(u) in [-1,1] lead to an open set in the usual sense of [0,1]. Then I need to say that the open subsets in [0,1] have open pre-images of f in X.

Trouble is, it's impossible to generalize an open subset. I think it's something like two half-open intervals at the end of the subset, with a union of open intervals in the middle?

Any help will be GREATLY appreciated.
Let $\displaystyle X,Y$ be $\displaystyle [0,1]$ with the usual and quotient topologies respectively. Show that $\displaystyle \text{id}:Y\to X$ is a homemorphism. Clearly it's bijective. To see it's continuous we recall that a map $\displaystyle f:Y\to Z$ is continuous if and only if $\displaystyle q\circ f:X\to Z$ is continuous. Thus, taking $\displaystyle Z=X$, $\displaystyle q=x^2$, and $\displaystyle f=\text{id}$ we see that $\displaystyle q\circf:X\to X:x\mapsto x^2$ is continuous because it's continuous with the usual topology. Try showing the other direction.

3. Wow thanks for the info, Drexel. So to show they're both equal it's enough to show that the identity function between both sets is a bijection, and prove continuity both ways?

If $\displaystyle q\circ f:X\to Z$ is continuous, this implies that for $\displaystyle U$ open in $\displaystyle Z$, $\displaystyle (q\circ f)^-^1$ open in X.
Since $\displaystyle (q\circ f)^-^1$ $\displaystyle = f^-^1(q^-^1(U))$...

If $\displaystyle = f^-^1(q^-^1(U))$ is open in X then this implies that $\displaystyle f^-^1$ must be open in X (somewhat clear). Therefore the pre-image of $\displaystyle f$ is open, which confirms the continuity.

Will edit later.

4. Originally Posted by Throughpoint
Wow thanks for the info, Drexel. So to show they're both equal it's enough to show that the identity function between both sets is a bijection, and prove continuity both ways?

If $\displaystyle q\circ f:X\to Z$ is continuous, this implies that for $\displaystyle U$ open in $\displaystyle Z$, $\displaystyle (q\circ f)^-^1$ open in X.
Since $\displaystyle (q\circ f)^-^1$ $\displaystyle = f(q(U)^-^1)$...

Will edit later.
Yes. And go ahead. I'll check it then.

5. Done I think?

Nah that's not right. I don't know where to go from the fact that $\displaystyle f^-^1(q^-^1(U))$ is open in X..