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

**Throughpoint** · November 5th 2010, 10:47 AM

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.

**Drexel28** · November 5th 2010, 11:13 AM

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

**Throughpoint** · November 5th 2010, 01:28 PM

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

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

Will edit later.

**Drexel28** · November 5th 2010, 02:40 PM

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

Will edit later.

Yes. And go ahead. I'll check it then.

**Throughpoint** · November 5th 2010, 04:44 PM

Done I think?

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

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

a topology such that closed sets form a topology

Topology homeomorphisms and quotient map question

Usual topology

How to show Topology and find basis

Topology Quotient Space

Search Tags