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 $e_b : E/R \to Y$.
$R$ 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!!