Let \(f:X\to Y\) and \(g:Y\to Z\) be functions. Show that if \(f\) and \(g\) are bijective, then so is \(g\circ f\), and we have \((g\circ f)^{-1}=f^{-1}\circ g^{-1}\).
- \(\bullet\) Show that \(g\circ f\) is bijective, when
- – C1: \(f\) is bijective
- – C2: \(g\) is bijective
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\(\Vdash\) { Proof by showing that \(g\circ f\) is injective and surjective }
- \(\bullet\) Show that \(g\circ f\) is injective
- \(\Vdash\) \((g\circ f) (x)=(g\circ f) (x')\)
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\(\Rightarrow\) { \(g\) is injective by C2 and Definition 3.3.14 }
- \(f(x)=f(x')\)
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\(\Rightarrow\) { \(f\) is injective by C1 and Definition 3.3.14 }
- \(x=x'\)
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\(\Rightarrow\) { Definition 3.3.14 }
- \(g\circ f\) is injective
- \(\square\)
- \(\bullet\) Show that \(g\circ f\) is surjective
- \(\Vdash\) \(g\) is surjective by C1
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\(\equiv\) { Definition 3.3.17 }
- \(\forall z\in Z,\exists y\in Y.~g(y)=z\)
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\(\Rightarrow\) { \(f\) is surjective by C2, Definition 3.3.17: }
- \(\forall z\in Z,\exists y\in Y,[\forall y\in Y,\exists x\in X.~f(x)=y)]. g(y) = z\)
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\(\equiv\) { \(\exists y\in Y,[\forall y\in Y, \exists x\in X. f(x)=y]. g(y)\Rightarrow \exists x. g(f(x))\) }
- \(\forall z\in Z,\exists x\in X.~g(f(x))=z\)
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\(\equiv\) { Definition 3.3.17 }
- \(g\circ f\) is surjective
- \(\square\)
- \(\square\)