Surjective Proof

Proof Surjective (or Onto 满射 )

Let f: R² -> R²
(x,y) -> (x+1, y+2)
Prove f onto?

Proof:
By defn of f, f(R²) ⊆ R²
To be onto: f(R²) = R²
=> we must prove:
f(R²) ⊇ R²

∀ (x’, y’) ∈ R² ...(1)
We must show that
(x’, y’) ∈ f(R²),
∃ (x, y) in domian R²
s.t. f(x,y)=(x’,y’)
x’=x+1, y’=y+2
x=x’-1, y=y’-2
With these values (x,y)∈R² and f(x,y)=(x’,y’) as required,
=> (x’, y’) ∈ f(R²)...(2)
Thus R² ⊆ f(R²)

f(R²) ⊆ R² and R² ⊆ f(R²)
=> f(R²) = R²= codomain
=> f is onto
[QED]