Exercise · ZW-COMPOSITION-AND-INVERSE-017 · ★★☆ Intermediate
If \(f\colon A\to B\), why is \(f\circ\mathrm{id}_A=f\)?
Choose one option.
Show answer
- B. Because \(\mathrm{id}_A\) leaves each \(x\) fixed, so \(f(\mathrm{id}_A(x))=f(x)\)
Explanation
Since \(\mathrm{id}_A(x)=x\), applying it before \(f\) changes nothing, so \(f\circ\mathrm{id}_A\) agrees with \(f\) at every point. No bijectivity of \(f\) is needed.
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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