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

• Nov 5th 2010, 10:47 AM
Throughpoint
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.
• Nov 5th 2010, 11:13 AM
Drexel28
Quote:

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.
• Nov 5th 2010, 01:28 PM
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 $\displaystyle (q\circ f)^-^1$ $\displaystyle = f^-^1(q^-^1(U))$...

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.
• Nov 5th 2010, 02:40 PM
Drexel28
Quote:

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.
• Nov 5th 2010, 04:44 PM
Throughpoint
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..