Let f:A→ B be surjective. Using the axiom of choice, which function g:B→ A can be constructed?

A g such that f(g(b))=b for all b∈ B.

A g such that g(f(a))=a for all a∈ A (an exact inverse).

A g with g(b)=f(b) for every b∈ B.

None, because f is not injective.

Exercise · ZW-AXIOM-OF-CHOICE-008 · ★★☆ Intermediate

Let \(f:A\to B\) be surjective. Using the axiom of choice, which function \(g:B\to A\) can be constructed?

Choose one option.

Show answer

D. A \(g\) such that \(f(g(b))=b\) for all \(b\in B\).

Explanation

Since \(f\) is surjective, each fiber \(f^{-1}(b)\) is nonempty, and choosing one point in each gives a section \(g\) with \(f\circ g=\mathrm{id}_B\). Demanding an exact inverse only works when \(f\) is bijective.

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Review the concept: Axiom of choice

Statement that every family of nonempty sets admits a choice function \(f\) that picks one element from each set in the family.

\[\forall X\,(\emptyset\notin X \rightarrow \exists f\,\forall A\in X\,(f(A)\in A))\]

Let \(f:A\to B\) be surjective. Using the axiom of choice, which function \(g:B\to A\) can be constructed?

Explanation

Review the concept: Axiom of choice

More practice problems: axiom of choice