Finding a useful denial of a injective function and a surjective function

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Homework Statement

Find the useful denial of a injective function and a surjective function.

Homework Equations

The Attempt at a Solution

I know a one to one function is (∀x₁,x₂ ∈ X)(x₁≠x₂ ⇒ f(x₁) ≠ f(x₂)). So would the useful denial be (∃x₁,x₂ ∈ X)(x₁ ≠ x₂ ∧ f(x₁) = f(x₂))?

I know a onto function is (∀y ∈ Y)(∃x ∈ X)(y=f(x)). So would the useful denial be (∃y ∈ Y)(∀x ∈ X)(y≠f(x))?

Thank you.

 

[fresh_42]

Homework Statement

Find the useful denial of a injective function and a surjective function.

Homework Equations

The Attempt at a Solution

I know a one to one function is (∀x₁,x₂ ∈ X)(x₁≠x₂ ⇒ f(x₁) ≠ f(x₂)). So would the useful denial be (∃x₁,x₂ ∈ X)(x₁ ≠ x₂ ∧ f(x₁) = f(x₂))?

I know a onto function is (∀y ∈ Y)(∃x ∈ X)(y=f(x)). So would the useful denial be (∃y ∈ Y)(∀x ∈ X)(y≠f(x))?

Thank you.

Both is correct.

 

[ver_mathstats]

Both is correct.

Thank you.

 

[fresh_42]

Injective means

cannot happen, and surjective means

$y$ cannot exist.