Topology with quotient topology...
mina_world
Topology space (X, T).
Let R be an equivalence relation on the nonempty set X.
Let X/R be the set of all equivalence class.
Define that f : X -> X/R , f(x) = x/R
(x/R : the equivalence class determined by the element x.)
Give X/R the quotient topology. (X/R, T')
(In fact, f is surjective.)
Prove or counter-example.
(1) If U in T' , then f^{-1}(U) in T.
(2) If C is a compact subset of X, then f(C) is a compact subset of X/R.
(3) If V in T, then f(V) in T'.
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Are you ready ? OK, let's go...
(1) By definition of quotient map.
(2) Let {V_i | i in I} be a open covering of f(C).
f^{-1}(V_i) is open in X.
Of course, C subset Union_i [f^{-1}(V_i)]
so, {f^{-1}(V_i) | i in I} is a open covering of C.
Since C is a compact subset of X,
there is a finite subcovering {f^{-1}(V_1), ... , f^{-1}(V_n)}
of {f^{-1}(V_i) | i in I}.
Since f is surjective, f[f^{-1}(V_i)] = V_i.
so, C subset f^{-1}(V_1) U ... U f^{-1}(V_n)}
==> f(C) subset V_1 U ... U V_n.
so, there is a finite subcovering of {V_i | i in I}.
so, f(C) is a compact subset of X/R.
(3) Sorry. I need your advice.
Rotwang
> Hello teacher~
>
> Topology space (X, T).
>
> Let R be an equivalence relation on the nonempty set X.
> Let X/R be the set of all equivalence class.
>
> Define that f : X -> X/R , f(x) = x/R
> (x/R : the equivalence class determined by the element x.)
>
> Give X/R the quotient topology. (X/R, T')
> (In fact, f is surjective.)
>
> Prove or counter-example.
> (1) If U in T' , then f^{-1}(U) in T.
> (2) If C is a compact subset of X, then f(C) is a compact subset of X/R.
> (3) If V in T, then f(V) in T'.
>
>
> (3) Sorry. I need your advice.
Construct a counterexample as follows: let X be the unit interval
[0,1], and let the equivalence relation R consist only of pairs <x,x>
and also of the pairs <0,1> and <1,0>. IOW this equivalence relation
identifies the two ends of the interval. Then it is not hard to see
that the quotient topology on [0,1]/R makes it homeomorphic to the
circle. Now consider the image under f of the open set (1/2,1].
Gonçalo Rodrigues
<mina_...@hanmail.net> fed this fish to the penguins:
The proof is correct, but as it shows this as nothing to do with the
fact that the codomain is a quotient space. It just so happens that
the image of a compact set by a continuous functions is compact.
>(3) Sorry. I need your advice.
>
Hint: identify the two ends of the unit interval to obtain the circle.
Hope it helps, regards,
G. Rodrigues
mina_world
"Rotwang" <sg...@hotmail.co.uk> wrote in message
news:914d720b-8dac-4b30...@59g2000hsb.googlegroups.com...
More exact...
Lemma 1)
If f : X -> Y is surjective, continuous, open(or closed) map,
then f is quotient map.
Lemma 2)
Let f : X -> Y be surjective and quotient map.
Let xRx' when f(x) = f(x').
then R is an equivalence relation on X.
and X/R ~ Y by q : X -> X/R, q(x) = x/R.
-------------------------------------------------------
Let g : X -> Y, g(t) = (cos 2pi.t , sin 2pi.t) , X = [0,1].
so, Y = {(x, y) | x^2 + y^2 = 1}
and g is surjective, continuous, closed map.
so, g is quotient map by lemma 1)
Define that R is a relation on X with tRt' <=> f(t) = f(t').
so, R is an equivalence relation on X. (easy to show.)
so, X/R = {{t} | 0 < t < 1} U {{0, 1}} (Because, f(0) = f(1))
Let f : X -> X/R , f(x) = x/R.
so, X/R ~ Y by lemma 2). (homeomorphism).
Since g : X -> Y is a quotient map,
I can give the quotient topology of X/R by quotient topology of Y.
and
Since g : X -> Y is not open map,
(consider the image under f of the open set (1/2,1].)
so, f : X -> X/R is also not open map.
mina_world
"mina_world" <mina_...@hanmail.net> wrote in message
news:ftcd3i$vri$1...@localhost.localdomain...
pf) omission.
> Lemma 2)
> Let f : X -> Y be surjective and quotient map.
> Let xRx' when f(x) = f(x').
>
> then R is an equivalence relation on X.
> and X/R ~ Y by q : X -> X/R, q(x) = x/R.
pf)
R is an equivalence relation on X. (easy to show.)
Let f(x) = y.
so, f^{-1}(y) = x/R
so, X/R = {f^{-1}(y) | y in Y}
Let h : Y -> X/R , h(y) = f^{-1}(y).
If y_1 =/= y_2, then f^{-1}(y_1) =/= f^{-1}(y_2).
so, h is injective and surjective.
NOW, I must show that h, h^{-1} are continuous. so, X/R ~ Y.
Let q : X -> X/R, q(x) = x/R.
Since q is surjective, I can give the quotient topology of X/R.
Now, suppose that q is a quotient map.
Namely,
f : X -> Y
h : Y -> X/R
q : X -> X/R
Since q = hof, f = h^{-1}oq,
For open G in X/R, q^{-1}[G] = f^{-1}[h^{-1}[G]] is open in X.
Since f is a quotient map, h^{-1}[G] must be open in Y.
Similar~
For open V in Y, f^{-1}[V] = q^{-1}[h(V)] is open in X.
Since q is a quotient map, h(V) must be open in X/R.
END.