Let \(f:X\to Y\) be a bijective function, and let \(f^{-1}:Y\to X\) be its inverse. Let \(V\) be any subset of \(Y\). Prove that the forward image of \(V\) under \(f^{-1}\) is the same set as the inverse image of \(V\) under \(f\); thus the fact that both sets are denoted by \(f^{-1}(V)\) will not lead to any inconsistency.
- \(\bullet\) Show that Forward image of \(V\) under \(f^{-1}\) = Inverse image of \(V\) under \(f\)
- \(\Vdash\) Forward image of \(V\) under \(f^{-1}\)
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\(=\) { Definition 3.4.1 }
- \(\{f^{-1}(y):y\in V\}\)
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\(=\) { \(f\) is bijective \(\Rightarrow f^{-1}\) is surjective
\(\quad~~\) Definition 3.3.17: \(\forall x\in X, \exists y\in Y, f^{-1}(y)=x\) }- \(\{x\in X:(x=f^{-1}(y))\land (y\in V)\}\)
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\(=\) { \(f\) is bijective \(\Rightarrow f\) is surjective
\(\quad~~\) Definition 3.3.17: \(\forall y\in V, \exists x\in X, f(x)=y\) }- \(\{x\in X:f(x)\in V\}\)
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\(=\) { Definition 3.4.4 }
- Inverse image of \(V\) under \(f\)
- \(\square\)
- \(\bullet\) Show that Forward image of \(V\) under \(f^{-1}\) = Inverse image of \(V\) under \(f\)
- \(\Vdash\) Forward image of \(V\) under \(f^{-1}\)
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\(=\) { Definition 3.4.1 }
- \(\{f^{-1}(y):y\in V\}\)
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\(=\) { \(f\) is bijective \(\Rightarrow f^{-1}\) is surjective
\(\quad~~\) Definition 3.3.17: \(\forall f^{-1}(y)\in X, \exists y'\in Y, f^{-1}(y')=f^{-1}(y)\) }- \(\{f^{-1}(y)\in X:(\exists y'\in Y, f^{-1}(y')=f^{-1}(y))\land (y\in V)\}\)
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\(=\) { \(f\) is surjective \(\Rightarrow\) \(\forall y', \exists x'\in X, f(x')=y'\) }
- \(\{f^{-1}(y)\in X:(\exists y'\in Y, \exists x'\in X, f^{-1}(f(x'))=f^{-1}(y))\land (y\in V)\}\)
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\(=\) { \(f\) is bijective \(\Rightarrow\) \(f^{-1}(f(x'))=x'\) }
- \(\{f^{-1}(y)\in X:(\exists x'\in X, x'=f^{-1}(y))\land(y\in V)\}\)
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\(=\) { \(f^{-1}(y)=x'\) }
- \(\{x'\in X:(x'=f^{-1}(y))\land(y\in V)\}\)
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\(=\) { \(f\) is bijective \(\Rightarrow\) \(x'=f^{-1}(y)\iff f(x')=y\) }
- \(\{x'\in X:(f(x')=y)\land (y\in V)\}\)
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\(=\) { \(f\) is bijective \(\Rightarrow f\) is surjective
\(\quad~~\) Definition 3.3.17: \(\forall y\in V, \exists x\in X, f(x)=y\) }- \(\{x'\in X:\exists x\in X, (f(x')=f(x))\land (f(x)\in V)\}\)
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\(=\) { \(\exists x\in X, (f(x')=f(x))\land (f(x)\in V)\Rightarrow f(x')\in V\) }
- \(\{x'\in X:f(x')\in V\}\)
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\(=\) { Definition 3.4.4 }
- Inverse image of \(V\) under \(f\)
- \(\square\)