# quotient map

• Apr 23rd 2010, 03:15 PM
rain07
quotient map
Recall that the integer part (or integral part) of a real number x is the unique integer n ∈ Z such that n ≤ x < n + 1. We denote it by I(x).
On R we deﬁne the relation xRy ⇔ I(x) = I(y).
(a) Let p : R → R/R be the quotient map, let R/R be endowed with the quotient topology, and let U be an open set in R/R. Prove that if n ∈ Z is such that p(n) ∈ U then p(n − 1) ∈ U .
(b) Deduce that the open sets in R/R are ∅, R/R and the image sets p(−∞, n] , where n ∈ Z.
(c) Consider the map I : R → Z, x ↦→ I(x) . Is the map I continuous (when Z is endowed with the induced topology) ?
Prove that I deﬁnes a bijection I : R/R → Z. What is the topology on Z making I a homeomorphism ?

I can see that p(n) ∈ U, let U =(x,y), then n ∈ (x,y), p(n-1) = n-1 ∈(x-1,y-1), p^-1(U) ∈(x-1, y), but I'm not sure if it's correct and how to proceed with the rest of the questions. Any input is appreciated!
• Apr 23rd 2010, 03:44 PM
Drexel28
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Originally Posted by rain07
Recall that the integer part (or integral part) of a real number x is the unique integer n ∈ Z such that n ≤ x < n + 1. We denote it by I(x).
On R we deﬁne the relation xRy ⇔ I(x) = I(y).
(a) Let p : R → R/R be the quotient map, let R/R be endowed with the quotient topology, and let U be an open set in R/R. Prove that if n ∈ Z is such that p(n) ∈ U then p(n − 1) ∈ U .

I'm assuming $\displaystyle \mathbb{R}$ has the usual topology. So, we know that $\displaystyle n\in p^{-1}(U)$, right? But, since $\displaystyle p^{-1}(U)$ is open there exists some neighborhood $\displaystyle V$ of $\displaystyle n$ such that $\displaystyle V\subseteq p^{-1}(U)$. But, whatever neighborhood $\displaystyle V$ is it hits points strictly less than $\displaystyle n$ and so points such that $\displaystyle I(x)=n-1$. So, let $\displaystyle x$ be one of those points. Then, $\displaystyle x\in V\subseteq p^{-1}(U)\implies p(x)=p(n-1)\in p\left(p^{-1}(U)\right)=U$ the last part gotten since every quotient map is surjective.
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(b) Deduce that the open sets in R/R are ∅, R/R and the image sets p(−∞, n] , where n ∈ Z.
You can do this.

Quote:

(c) Consider the map I : R → Z, x ↦→ I(x) . Is the map I continuous (when Z is endowed with the induced topology) ?
Prove that I deﬁnes a bijection I : R/R → Z. What is the topology on Z making I a homeomorphism ?
Of course the map $\displaystyle I$ isn't continuous. For, given any $\displaystyle z\in\mathbb{Z}$ we have that $\displaystyle \left(z-\tfrac{1}{2},z+\tfrac{1}{2}\right)\cap\mathbb{Z}=\ {z\}$ is open and so $\displaystyle \mathbb{Z}$ with the subspace topology is discrete. And so, if $\displaystyle I$ were continuous then $\displaystyle I^{-1}(\{0\})$ should be open, but $\displaystyle I^{-1}(\{0\})=[0,1)$

I think you have your map a little screwed up. So you're saying that $\displaystyle I:\mathbb{R}/\sim\to\mathbb{Z}:p(x)\mapsto I(x)$, right? It doesn't go without saying that the map is well defined. But, if $\displaystyle p(x)=p(y)$ then $\displaystyle x=n+\varepsilon$ for some $\displaystyle 0\leqslant \varepsilon<1$ and $\displaystyle y=n+\delta,\text{ }0\leqslant\delta<1$ from where it follows that $\displaystyle I(p(x))=I(n+\varepsilon)=n=I(y+\delta)=I(p(y))$. Now, to see it's an injection suppose that $\displaystyle I(x)=I(y)$ then $\displaystyle x=n+\varepsilon$ and $\displaystyle y=n+\delta$ just as before and so $\displaystyle p(x)=p(y)$. To see it's a surjection you merely must note that $\displaystyle I(p(z))=z,\text{ }\forall z\in\mathbb{Z}$.

I would say you need to give it the quotient topology determined by $\displaystyle \overset{\sim}{I}:\mathbb{R}\to\mathbb{Z}:x\mapsto I(x)$

So, to see now that $\displaystyle I:\mathbb{R}/\sim\to\mathbb{Z}$ is continuous we need merely note that a mapping out of a quotient space is continuous if and only if it's composition with the definining quotient map is continuous. But, $\displaystyle I\circ p:\mathbb{R}\to\mathbb{Z}:x\overset{p}{\longmapsto }p(x)\overset{I}{\mapsto}I(x)$ and of course this is continuous, it is the quotient map used to define the topology on $\displaystyle \mathbb{R}/\sim$!

So, why is it's inverse $\displaystyle I^{-1}:\mathbb{Z}\to\mathbb{R}/\sim:x\mapsto z\mapsto p(z)$ continuous? Well, let's compose it with the quotient map defined in the making of $\displaystyle \mathbb{R}/\sim$. We get....I'll let you finish this.