Talk:Quotient space (topology)

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

message: 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)
thread_title:  im ''&fnof;'' homeomorphic to X/~ 

message: 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)
thread_title:  Characterization of quotient maps 

message: 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)
thread_title:  Easy Example 

message: 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)
thread_title:  Wedge Sum 

message: 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)
thread_title: Definition of Quotient map

message: 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)
thread_title:  Quotient map 

message: In Properties, the second picture seems over-large.
Darcourse (talk) 10:55, 19 February 2022 (UTC)
thread_title:  Different picture sizes 

message: 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) \ne 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)
thread_title:  Sloppy use of preimage functions 

message: 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  until a consensus is reached.  1234qwer1234qwer4 11:26, 18 April 2023 (UTC)
thread_title:  "[[:Quotient map]]" listed at [[Wikipedia:Redirects for discussion|Redirects for discussion]]