Exercise · ZW-COMPOSITION-AND-INVERSE-009 · ★★★ Advanced

Let \(f\colon A\to B\) and \(h\colon B\to A\) satisfy \(h\circ f=\mathrm{id}_A\). What can be said about \(f\)?

Choose one option.

Show answer
  • B. \(f\) is injective

Explanation

If \(f(a)=f(a')\), applying \(h\) gives \(a=h(f(a))=h(f(a'))=a'\), so \(f\) is injective. It need not be surjective: if \(B\) is larger than \(A\), \(f\) cannot cover it.

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Review the concept: Composition and inverse

The composition \(g\circ f\) applies \(f\) first and then \(g\). The inverse \(f^{-1}\) exists when \(f\) is bijective and undoes its action.

\[(g\circ f)(x)=g(f(x)), \quad (g\circ f)^{-1}=f^{-1}\circ g^{-1}\]

More practice problems: composition and inverse

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