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For the following question:

Let $X$ be the Cartesian plane $\mathbb{R}^2$. Equip $X$ with the topology induced by the sub-basis
? $$E_{{x},{y},{\epsilon}} =\{(s,t)\in X: s=x, y-\epsilon<t<y+\epsilon\}$$
for $x,y\in \mathbb{R}, \epsilon > 0$. Define $T:X\rightarrow \mathbb{R}$ by $T(x,y)=x$. What quotient topology is induced on $\mathbb{R}$ by this mapping?

I think the quotient topology is the euclidean topology. I am not sure. But to satisfied the definition of quotient topology, where the definition is as follows:

Definition: Let $(X,U)$ be a topological space and let $Y$ be any set. Suppose that $f:X\rightarrow Y$ is a surjective mapping. Then the collection of subsets $$\tau_f=\{{G\subset Y:f^{-1}(G) \text{ is open in } X\}}$$ is a topology on $Y$ called the quotient topology induced on $Y$ by $f$. When $Y$ is endowed with such a quotient topology, then it is called a quotient space of $X$, and the inducing map $f$ is called a quotient map.

Here, we let the mapping $T$ to stand for $f$. I need to do two things, one, for any open $G \subset Y$, $T^{-1}(G)$ is open in $X$. Two, I need to show that the open sets $G$ collectively form a topology.
The thing is, the domain of $T$ is $X \times Y$. So would the inverse image of $T$ consist of the form $U_{x} \times Y$. Because from the topology form from the sub basis:

$E_{{x},{y},{\epsilon}} =\{(s,t)\in X: s=x, y-\epsilon<t<y+\epsilon\}$

I am not sure if the open sets are of the form $U_{x} \times Y$ or is it of the form $X \times G_{y}$. I know that $G_{y}=(y-\epsilon, y+\epsilon)$

Seth
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    I believe in your definition of $E_{x,y,\epsilon}$ it should read '${ (s,t) \in X ...}$', right? Otherwise, I would not know where $s$ and $t$ are coming from. – Matthias Klupsch Jan 07 '20 at 14:45
  • An open set is a union of ${x} \times G_{y, \varepsilon}$'s – Berci Jan 07 '20 at 14:54
  • @MatthiasKlupsch1 thank you for pointing that out. I have corrected it. – Seth Jan 07 '20 at 16:16
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    Why do you think the quotient topology is Euclidean? If you take a singleton in $\mathbb{R}$ say ${x}$, then, $T^{-1}(x) = {(x, y) : y \in \mathbb{R}}$ which can be written as a union of the subbasis elements, for example as: $\bigcup_{y\in \mathbb{Z}} E_{x, y, 1}$. – feynhat Jan 07 '20 at 16:38
  • @FabioLucchini. I d.k. why you deleted your A, as it's correct. The quotient topology on $\Bbb R$ is the strongest topology such that $T$ is continuous. – DanielWainfleet Jan 08 '20 at 01:17
  • @feynhat I found the solution/hint but I don't understand it. It states: "A subset $U$ of $\mathbb{R}$ pulls back to an open set in $X$ if and only if it is an open interval. Hence the quotient topology is the usual Euclidean topology". I don't know the meaning of the use of the phrase "pulls back". Also, when in the case of quotient topology, I am not sure if being the strongest topology is the same as being the finest topology in the sense of having the most open sets like in the case of the box vs the product topology. – Seth Jan 09 '20 at 05:44
  • @SethMai In this context, I suppose, "pull back" means inverse image. It is not clear to me why only intervals will have inverse image as open sets. As I commented earlier, every singleton has a inverse image as open sets, so the topology should be discrete. Can you link me to the problem/solution? – feynhat Jan 10 '20 at 05:38
  • @feynhat https://books.google.ca/books?id=LUhabKjfQZYC&pg=PA95&source=gbs_toc_r&cad=4#v=onepage&q&f=false chapter 2 question 14, there is hint/solution at the back of book. There are so many possibilities what the quotient topology could be, I guessed it was euclidean and not product or lower/upper limit one. I actually don't know how to show if it is the strongest topology indpendent of being a quotient topology. – Seth Jan 10 '20 at 05:48
  • @SethMai I am not convinced by the answer given in the book. The inverse image of a singleton is a line parallel to y-axis, which is an open set in the given topology (you can write this line as a union of sub-basis elements). – feynhat Jan 10 '20 at 12:07
  • @SethMai Regarding your question about quotient topology being the strongest topology (or largest or finest, these words refer to the same thing, ie., the maximal element in the poset of topologies satisfying some condition): note that according to your definition, in the quotient topology $G$ is open iff. $q^{-1}(G)$ is open. We claim that this is the largest topology such that $q$ is continuous. If not, then there is a set $G'$ (not in quotient topology) such that $q^{-1}(G)$ is open, which contradicts the definition of quotient topology. – feynhat Jan 10 '20 at 12:14
  • Coming back to our case, since every singleton has open inverse image. Every singleton will be open in the quotient topology. So, every set will be open. Or, in other words, the quotient topology is discrete. – feynhat Jan 10 '20 at 12:17
  • @feynhat are strong and weak topologies the same as initial and final topologies in meaning? Thanks – Seth Jan 11 '20 at 22:10

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