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Let $E$ be the set of all functions from a set $X$ into a set $Y$. Let $b \in X$ and let $R$ be the subset of $E \times E$ consisting of those pairs $(f,g)$ such that $f(b) = g(b)$. Prove that $R$ is an equivalence relation. Define a bijection $e_b : E/R \to Y$.

My work:

$R$ is an equivalence relation is clear. Define $e_b :E/R \to Y$ as $[f] \mapsto f(b)$. Then $e_b$ is injective: Assume $f(b) = g(b)$. Then by definition, $[f] = [g]$. Also, $e_b$ is surjective: Let $y \in Y$. Then $f(x) = y$ for all $x$ is a function mapping $b \to y$. Hence $e_b ([f]) = f(b) = y$.

Please correct me if I'm wrong. Many thank you!!

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It looks correct. – Adeel Feb 7 '13 at 10:27
@Adeel Thank you for correcting me!! – Visitor Feb 7 '13 at 13:18
Then is $R$ the kernel of the function $P_b:E\rightarrow Y$ defined as $f \mapsto P_b(f)=f(b)$? so we have that $e_b([f]_R)=P_x(f)$ – MphLee Mar 20 '14 at 20:10

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