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Let $f:X \to Y$ be a surjective function from a set $X$ onto a set $Y$. Let $R$ be the subset of $X \times X$ consisting of those pairs $(x,x')$ such that $f(x) =f(x')$. Prove that $R$ is an equivalence relation. Let $\pi : X \to X/R$ be the projection. Verify that, if $\alpha \in X/R$ is an equivalence class, to define $F(\alpha) = f(a)$ whenever $\alpha = \pi (a)$ establishes a well-defined function $F:X/R \to Y$ that is surjective and injective.

My work: $R$ is an equivalence relation is clear.

$F$ is well-defined: Let $\alpha = \pi (a) = \pi (b) = \beta$. Then $F(\alpha ) = f(a) = f(b) = F(\beta)$.

$F$ is injective: Let $F(\alpha ) = f(a) = f(b) = F(\beta)$. Then $ \pi (a) = \pi (b) $ and hence $\alpha = \beta$.

$F$ is surjective: Let $y \in Y$. Since $f$ is surjective there is $x$ with $f(x) =y$. Hence $F(\pi(x)) = f(x) = y$.

Thank you for correcting me..

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This is correct, but it lacks some redaction "let $a$ and $b$ such that $F(\pi(a)) = F(\pi(b))$ then blabla". It is important to explain what are the variables you use and what are the assumptions on them. – Damien L Feb 7 '13 at 10:17
@DamienL Thank you!! – Visitor Feb 7 '13 at 10:20

Your work is correct. As was mentioned in the comments, your work could be improved readability-wise by giving some intuition behind them. For example, what is the big idea?

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