Exercise · ZW-SURJECTIVE-FUNCTION-004 · ★☆☆ Basic

If \(f\colon A\to B\) is not surjective, what holds?

Choose one option.

Show answer
  • B. There exists \(b\in B\) that is not the image of any \(a\in A\)

Explanation

The negation of «every element of \(B\) is an image» is «some \(b\) has no preimage». The option about pairs with equal images describes failing injectivity, which does not follow from failing surjectivity: \(n\mapsto 2n\) from \(\mathbb{Z}\) to \(\mathbb{Z}\) is not surjective but is injective.

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Review the concept: Surjective function

Function \(f\colon A\to B\) whose image is the whole codomain \(B\).

\[\forall b\in B\,\exists a\in A\,(f(a)=b)\]

More practice problems: surjective function

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