Proof
(related to Proposition: Functions Constitute Equivalence Relations)
 By hypothesis, $f:A\mapsto B$ is a function and the relation "$\sim$" defined by $x\sim y:\Leftrightarrow f(x)=f(y).$
 Clearly, $\sim$ is reflexive, since $f(x)=f(x)$ for every $x\in A.$
 Also, $\sim$ is symmetric, since $f(x)=f(y)$ if and only if $f(y)=f(x)$ for all $x,y\in A.$
 Finally, $\sim$ is transitive, since if $f(x)=f(y),$ and $f(y)=f(z)$, then $f(x)=f(z)$ for all corresponding $x,y,z\in A.$
 This proves that $\sim$ is an equivalence relation with the corresponding quotient set $A/_{f}$ of elements in $A$ having the same image in $B.$
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References
Bibliography
 Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994