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, $f:X\rightarrow Y$ gives rise to an equivalence relation, where $x_1\sim x_2$ if $f(x_2)=f(x_2)$ but this doesn't use the surjective part!

• Notice also that in such a case we can factor $f$ through $\pi:X\rightarrow X/\sim$ always to yield $\bar{f}$, and "distil" the information of $f$ into this new map, $\bar{f}$.

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

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Flesh out, check existing content then demote to grade F

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Something in Topological Manifolds (Lee) should cover this. Under Quotient topology

Statement

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

• For $x_1,x_2\in X$, we say $x_1\sim x_2$ if $f(x_1)=f(x_2)$

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

• See also: every equivalence relation gives rise to a surjective map, usually written $\pi:X\rightarrow X/\sim$ given by $\pi:x\mapsto[x]$ where $[x]:=\{y\in X\ \vert x\sim y\}$, an equivalence class

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.

Proof

Grade: C

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This is easy. Truly

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