Session 3D
Composition and inverses
45-75 min - work through lesson notes, practice, and the MCQ check.
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Session tips
- Finish the MCQ before marking complete.
- Check the answer outlines after trying on paper.
- Mark complete to update your certificate progress.
Lesson notes
- Function composition is defined by (f o g)(x) = f(g(x)). Composition is a way to build complex functions from simpler ones, and many proof properties behave nicely under composition. For instance, if f and g are injective, then f o g is injective, because equality of outputs forces equality step-by-step going backward through g and then through f. If f and g are surjective, then f o g is surjective, because you can hit any target y by first finding an input for f and then finding an input for g.
- An inverse function f^-1 exists exactly when f is bijective. The defining property is f^-1(f(x)) = x for all x in the domain and f(f^-1(y)) = y for all y in the codomain. Proving something is an inverse is usually done by verifying these two identities. Beginners often treat inverses as automatic algebraic "undoing," but that only works if the function is one-to-one and onto in the specified sets. Thinking set-theoretically about inverses prevents errors like claiming x^2 has an inverse on all real numbers.
Practice
- 1If f and g are injective, prove f o g is injective.
- 2If f and g are surjective, prove f o g is surjective.
- 3Give an example where f o g is bijective but f is not bijective (codomains differ).
Show answers / outlines
Answers
- 1f(g(a)) = f(g(b)) -> g(a) = g(b) -> a = b.
- 2Use surjectivity of f and g to hit any y.
Week quiz
- 1Prove: A subseteq B implies A union C subseteq B union C.
- 2Decide if f(x) = x^2 from R to [0, infinity) is surjective.
Show quiz answers
Quiz answers
- 1Use subset proof: take x in A union C and show x in B union C.
- 2Yes, it is surjective onto [0, infinity).
MCQ
If f and g are injective, then f o g is: