# Talk:Quotient space (topology)

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Field: Topology

Can anyone tell me why its called "quotient"?

## im f 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.

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