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Let $\sim$ be an equivalence relation on a set $X$. Also, there is a natural function $p:X\to \tilde X$ where $\tilde X$ is a set of all equivalence classes. (Equivalence classes are defined as, $[x]=\{y \in X |x\sim y\}$ where the equivalence relation is reflexive, symmetric, and transitive $\forall (x,y)$). This natural function $p$ is defined by $p(x)=[x]$. When is this function surjective and when is it injective?

My guess was that it was surjective from $x$ to some $k\in \mathbb{N}$ and injective in $\mathbb{N}$, but I am probably wrong.

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up vote 9 down vote accepted

It's always surjective, as every equivalence class contains at least one element. It will fail to be injective any time two different elements are equivalent to each other, as if $x\sim y$ then $[x]=[y]$. So it is only injective if the equivalence relation is that of equality.

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I upvoted this because I think it's a reasonable explanation. –  Casquibaldo Sep 27 '12 at 21:48
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It is always surjective, and it is injective exactly when $\sim$ is the relation of equality.

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