How do the examples f₁ through f₆ illustrate function properties?

"With X = {Harry, Ron, Hermione} and Y = {Dr. Pepper, Mt. Dew}, consider the function f₁: f₁(Harry) = Mt. Dew, f₁(Ron) = Mt. Dew, and f₁(Hermione) = Mt. Dew. This function is not injective, since more than one wizard maps to Mt. Dew. It is not surjective, since no wizard maps to Dr. Pepper, and it is not bijective because it must be both injective and surjective. For f₂, change Ron to map to Dr. Pepper instead: f₂(Harry) = Mt. Dew, f₂(Ron) = Dr. Pepper, and f₂(Hermione) = Mt. Dew. It is still not injective, since more than one wizard maps to Mt. Dew. It is now surjective, since every soft drink has at least one wizard mapping to it, but it is not bijective. Now add Pepsi and Barq’s Root Beer to Y, so that it has four elements: {Dr. Pepper, Mt. Dew, Pepsi, Barq’s Root Beer}. Consider f₃: f₃(Harry) = Pepsi, f₃(Ron) = Pepsi, and f₃(Hermione) = Mt. Dew. It is not injective, since more than one wizard maps to Pepsi, and it is not surjective, since no wizard maps to Dr. Pepper or Barq’s. For f₄, change Ron to map to Dr. Pepper instead: f₄(Harry) = Pepsi, f₄(Ron) = Dr. Pepper, and f₄(Hermione) = Mt. Dew. It is still not surjective, but it is injective, since no drink has more than one wizard. Finally, add Neville to the mix. Let f₅ be: f₅(Harry) = Barq’s Root Beer, f₅(Ron) = Dr. Pepper, f₅(Hermione) = Mt. Dew, and f₅(Neville) = Dr. Pepper. It is not injective, since Dr. Pepper has two wizards, and it is not surjective, since Pepsi has none. However, with one small change, f₆ is defined by f₆(Harry) = Barq’s Root Beer, f₆(Ron) = Pepsi, f₆(Hermione) = Mt. Dew, and f₆(Neville) = Dr. Pepper. This last function is injective, surjective, and bijective: every wizard gets their own soft drink, every soft drink gets its own wizard, and no soft drinks or wizards are left out. The only way to get a bijection is for the domain and codomain to be the same size, although that alone does not guarantee a bijection, as f₅ demonstrates. If they are the same size, injectivity and surjectivity go hand-in-hand: violate one, and you violate the other; uphold one, and you uphold the other."