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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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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.2012-09-27
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It is always surjective, and it is injective exactly when $\sim$ is the relation of equality.