Exercise · ZW-COMPOSITION-AND-INVERSE-014 · ★★★ Advanced
If \(g\circ f\) is injective, what can be said about \(g\)?
Choose one option.
A \(g\) need not be injective B \(g\) is constant C \(g\) is the identity D \(g\) is injective
Show answer A. \(g\) need not be injective
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Explanation If \(g\) identifies two points, \(g\circ f\) can still be injective when \(f\) never reaches them. With \(A=\{1\}\), \(B=\{1,2\}\), \(C=\{a\}\) and \(g\) constant, \(g\circ f\) is injective.
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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