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
Exercise · ZW-INJECTIVE-FUNCTION-009 · ★★☆ Intermediate
Choose one option.
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.
Function \(f\colon A\to B\) in which distinct elements of \(A\) have distinct images in \(B\).