Exercise · ZW-COMPOSITION-AND-INVERSE-014 · ★★★ Advanced

If \(g\circ f\) is injective, what can be said about \(g\)?

Choose one option.

Show answer
  • A. \(g\) need not be injective

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.

Study Set theory in ZWMath →

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

← ZW-COMPOSITION-AND-INVERSE-013 ZW-COMPOSITION-AND-INVERSE-015 →