Talk:Quotient space (topology)

Name

Can anyone tell me why its called "quotient"? — Preceding unsigned comment added by Jordanl122 (talk • contribs) 00:13, 27 May 2006 (UTC)

im ƒ homeomorphic to X/~

I made a correction there--if f is not an open map, there is a continuous bijection from X/~ to im f, but it is not a homeomorphism.--Todd 15:07, 19 July 2006 (UTC)

You're right in thinking that a homeomorphism needs to be an open map. However, in the text the topology of Y is defined as the finest topology that makes f continuous: V is open in Y if and only if it's preimage under f is open in X. A topology on Y wasn't assumed; it was constructed. Originally, Y was only assumed to be a set. The construction not only makes f continuous, it also makes it open. I think the original text was right so unless I hear back from you soon, I'm going to change the article back. (Perhaps I'll try to clarify this point.) Lunch 18:38, 19 July 2006 (UTC)

Quotient maps aren't always open maps. However, the natural quotient map taking a space to an orbit space is an open map. 76.21.73.242 (talk) 22:32, 5 April 2008 (UTC)

Characterization of quotient maps

I made a correction in the following statement (change from "characterized by" to "characterized among surjective maps by"):

Quotient maps q : X → Y are characterized by the following property: if Z is any topological space and f : Y → Z is any function, then f is continuous if and only if fq is continuous.

Suppose the property holds for a map $q:X\to Y$. While q being continuous and $V\subseteq Y$ being open iff $q^{-1}(V)$ is open are quite easy to prove, I believe we cannot show q is onto. For suppose it isn't and on Y we have the strongest topology making q continuous (so on Y\q(X) the topology is discrete). Then the property holds ($f^{-1}(U)$ is open iff $f^{-1}(U)\cap q(X)$ is open iff $q^{-1}(f^{-1}(U))=(fq)^{-1}(U)$ is open, but $q$ is not a quotient map (since it is not surjective). --87.205.250.125 (talk) 21:56, 8 August 2008 (UTC)

Easy Example

Since "it is easy to construct examples of quotient maps which are neither open nor closed," it is surely easy to provide such an example! Could someone provide two simple ones? F. G. Dorais (talk) 15:10, 20 May 2010 (UTC)

A simple example would be the quotient map q : R -> X such that q respects the equivalence relation x~ y iff x >0 and y>0 or x<=0 and y<=0. Then X is a two point set (a,b) where q^{-1)(a)=(\infty,0] and q^{-1}(b)=(0,\infty) so a is closed and b is open. Then q is not closed because q([1,2]) is open and it's not open because q( (-2,-1) ) is closed. —Preceding unsigned comment added by 74.219.234.106 (talk) 11:43, 3 December 2010 (UTC)

Wedge Sum

In the article, it says that gluing two points together in a space is the wedge sum. But the article on wedge sum says that the wedge sum is taking the disjoint union of two sets and gluing them together at one point from each. So there seems to be a mismatch in terminology. — Preceding unsigned comment added by 136.159.160.241 (talk) 18:34, 12 September 2012 (UTC)

Example order

I think we confuse people (me) by putting out the example of the orbit space R/Z before explaining it. If you don't know what's going on, it doesn't fit the definition of quotient space thus far explained, and if you're a person who'll stick there and try to puzzle it out rather than read past it, you lose minutes of your life and gain some frustration for no reason.

Definition of Quotient map

Is there any reason to include the word 'continuous' in the definition of Quotient Maps? Doesn't the statement, U belonging to Y is open in Y, iff f^{-1}(U) is open in X , include the definition of continuity in the only if part?

Aritrop (talk) 18:36, 12 March 2015 (UTC)

173.25.54.191 (talk) 08:44, 7 October 2013 (UTC)

Quotient map

This section is plagued with redundancy. Needs to be edited and simplified. — Preceding unsigned comment added by RutiWinkler (talk • contribs) 22:49, 25 December 2021 (UTC)

Different picture sizes

In Properties, the second picture seems over-large.

Darcourse (talk) 10:55, 19 February 2022 (UTC)

Sloppy use of preimage functions

This page doesn't bother to point out that $f^{-1}$ is being used to mean the preimage under the mapping $f$, and that for nontrivial quotient maps, there isn't an inverse map since the original mapping is not injective. This gets authors in trouble. See for example the section "Quotient space of fibers characterization" which claims that for a quotient map $q$, $q(x)=q^{-1}(q(x))$ which is clearly nonsense. On the right side is the set of all elements in an equivalence class in the domain space, and on the left is a point in the range (quotient) space, which is the equivalence class itself. These aren't even objects in the same space. What is true is that $q(x)=q(q^{-1}(q(x)))$. That is, the image of $x$ under the quotient mapping is the image of all points in the equivalence class of $x$.

One more thing. The definition of hereditarily quotient mappings is interesting, but it seems to be more a statement about the equivalence relation than about the mapping. I would argue that an example is essential if this definition is to be retained. Here is one that I think works: On a space $X$ define the equivalence relation $x\sim y$ iff $x$ and $y$ belong to one connected component of the space. Now take $T$ to be a subset of a single component of $X$ that is itself not connected, and take $x$ and $y$ to be in different components of $T$. In $X$, $x\sim y$, however in $T$ this is no longer then case, so for the quotient map $f(x)=f(y)$, however $f|_{T}(x)\neq f|_{T}(y)$. If this isn't what this definition is driving at, then that is all the more reason for an illustrative example! Robroot (talk) 16:56, 16 October 2022 (UTC)

"Quotient map" listed at Redirects for discussion

The redirect Quotient map has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2023 April 18 § Quotient map until a consensus is reached. 1234qwer1234qwer4 11:26, 18 April 2023 (UTC)