Exercise · ZW-INJECTIVE-FUNCTION-009 · ★★☆ Intermediate

Is every injective function also surjective?

Choose one option.

Show answer
  • D. No; \(n\mapsto n+1\) from \(\mathbb{N}\) to \(\mathbb{N}\) is injective and not surjective

Explanation

With \(\mathbb{N}=\{1,2,\dots\}\), \(n+1=n'+1\) implies \(n=n'\), so it is injective; but \(1\) is not \(n+1\) for any \(n\in\mathbb{N}\). The other options are false: a finite domain is not enough (for example \(\{1\}\to\{1,2\}\)), and injective functions from \(\mathbb{N}\) to \(\mathbb{N}\) exist, such as the identity.

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Review the concept: Injective function

Function \(f\colon A\to B\) in which distinct elements of \(A\) have distinct images in \(B\).

\[\forall a_1,a_2\in A\,\bigl(f(a_1)=f(a_2)\Rightarrow a_1=a_2\bigr)\]

More practice problems: injective function

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