Exercise · ZW-COMPOSITION-AND-INVERSE-013 · ★★☆ Intermediate
Why does order matter in a composition?
Choose one option.
Show answer
- B. Because in general \(f\circ g\neq g\circ f\) and the domains must also match
Explanation
Changing the order usually changes the result, and \(g\circ f\) only exists if the codomain of \(f\) fits in the domain of \(g\).
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-001 What does it mean for fcolon A→ B to have an inverse f^-1colon B→ A?
- ZW-COMPOSITION-AND-INVERSE-002 If fcolon A→ B and gcolon B→ C, from which set to which set does g∘ f go?
- ZW-COMPOSITION-AND-INVERSE-003 What is the inverse of the identity id_A?
- ZW-COMPOSITION-AND-INVERSE-004 Let f=\(1,b),(2,c),(3,a)\ be a bijection from \1,2,3\ onto \a,b,c\. What is f^-1(a)?
- ZW-COMPOSITION-AND-INVERSE-005 Which identity does the inverse f^-1 of a bijection fcolon A→ B satisfy?
- ZW-COMPOSITION-AND-INVERSE-006 With f(x)=x+1 and g(x)=x^2 on ℝ, what is (g∘ f)(2)?
- ZW-COMPOSITION-AND-INVERSE-007 Is g∘ f=f∘ g always true?
- ZW-COMPOSITION-AND-INVERSE-008 If f and g are bijections, what is (g∘ f)^-1?
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