# Every surjective map gives rise to an equivalence relation

## Warning:

Warning:There is something wrong here, as the surjective property is never used! It is true though that every map, [ilmath]f:X\rightarrow Y[/ilmath] gives rise to an equivalence relation, where [ilmath]x_1\sim x_2[/ilmath] if [ilmath]f(x_2)=f(x_2)[/ilmath] but this doesn't use the surjective part!

• Notice also that in such a case we can factor [ilmath]f[/ilmath] through [ilmath]\pi:X\rightarrow X/\sim[/ilmath] always to yield [ilmath]\bar{f} [/ilmath], and "distil" the information of [ilmath]f[/ilmath] into this new map, [ilmath]\bar{f} [/ilmath].
This page is a dire page and is in desperate need of an update.
The message is:
Why was it even created?! Fix the surjective problem

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Flesh out, check existing content then demote to grade F
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
Something in Topological Manifolds (Lee) should cover this. Under Quotient topology

## Statement

Let [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be sets, and let [ilmath]f:X\rightarrow Y[/ilmath] be a surjective mapping. Then we can define an equivalence relation on [ilmath]X[/ilmath], [ilmath]\sim\subseteq X\times X[/ilmath], by:

• For [ilmath]x_1,x_2\in X[/ilmath], we say [ilmath]x_1\sim x_2[/ilmath] if [ilmath]f(x_1)=f(x_2)[/ilmath]

So we have [ilmath]\forall x_1,x_2\in X[x_1\sim x_2\iff f(x_1)=f(x_2)][/ilmath] (for the "and only if" part, see definitions and iff)

## Purpose

The quotient topology can be defined using a map to identify parts together, but also an equivalence relation can do the same job, this page is half way to showing them to be the same.