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.
A \(f\) is surjective B \(f\) is injective C \(f\) is constant D \(f\) is bijective
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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.
The composition \(g\circ f\) applies \(f\) first and then \(g\). The inverse \(f^{-1}\) exists when \(f\) is bijective and undoes its action.
\[(g\circ f)(x)=g(f(x)), \quad (g\circ f)^{-1}=f^{-1}\circ g^{-1}\]
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