Set of Mappings which map to Same Element induces Equivalence Relation

Theorem
Let $X$ and $Y$ be sets.

Let $E$ be the set of all mappings from $X$ to $Y$.

Let $b \in X$.

Let $\mathcal R \subseteq E \times E$ be the relation on $E$ defined as:
 * $\mathcal R := \set {\tuple {f, g} \in \mathcal R: \map f b = \map g b}$

Then $\mathcal R$ is an equivalence relation.

Proof
Checking in turn each of the criteria for equivalence:

Reflexivity

 * $\forall f \in E: \map f b = \map f b$

That is:
 * $\forall f \in E: \tuple {f, f} \in \mathcal R$

Thus $\mathcal R$ is seen to be reflexive.

Symmetry
Thus $\mathcal R$ is seen to be symmetric.

Transitivity
Let $f, g, h \in E$.

Thus $\mathcal R$ is seen to be transitive.

$\mathcal R$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.