If \(X\) is a subset of \(Y\) , let \(\imath_{X \to Y} : X → Y\) be the inclusion map from \(X\) to \(Y\) , denoted by mapping \(x \mapsto x\) for all \(x \in X\), i.e., \(\imath_{X \to Y} (x) := x\) for all \(x \in X\). The map \(\imath_{X \to X}\) is in particular called the identity map on \(X\).
(a) Show that if \(X \subseteq Y \subseteq Z\) then \(\imath_{Y \to Z} \circ \imath_{X \to Y} = \imath_{X \to Z}\) .
(b) Show that if \(f : A \to B\) is any function, then \(f = f \circ \imath_{A \to A} = \imath_{B \to B} \circ f\) .
(c) Show that, if \(f : A \to B\) is a bijective function, then \(f \circ f^{-1} = \imath_{B \to B}\) and \(f^{-1} \circ f = \imath_{A \to A}\).
(d) Show that if \(X\) and \(Y\) are disjoint sets, and \(f : X \to Z\) and \(g : Y → Z\) are functions, then there is a unique function \(h : X \cup Y \to Z\) such that \(h \circ \imath_{X→X \cup Y} = f\) and \(h \circ \imath_{Y \to X \cup Y} = g\).
- \(\bullet\) (a) Show that if \(X \subseteq Y \subseteq Z\) then \(\imath_{Y \to Z} \circ \imath_{X \to Y} = \imath_{X \to Z}\)
- \(-\) Let \(x\) be any element in \(X\)
- \(\Vdash\) \((\imath_{Y \to Z} \circ \imath_{X \to Y}) (x)\)
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\(=\) { Definition 3.3.10 : function composition }
- \((\imath_{Y \to Z}) (\imath_{X \to Y} (x))\)
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\(=\) { \(X \subseteq Y\) and definition of the inclusion map }
- \((\imath_{Y \to Z}) (x)\)
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\(=\) { \(Y \subseteq Z\) and definition of the inclusion map }
- \(x\)
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\(=\) { \(X \subseteq Z\) and definition of the inclusion map }
- \((\imath_{X \to Z}) (x)\)
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\(\vdash\) { Definition 3.3.7 (Equality of functions) }
- \(\imath_{Y \to Z} \circ \imath_{X \to Y} = \imath_{X \to Z}\)
- \(\square\)
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\(\bullet\) (b) Show that if \(f : A \to B\) is any function, then \(f = f \circ \imath_{A \to A} = \imath_{B \to B} \circ f\) .
- \(-\) Let \(a\) be any element in \(A\)
- \(\Vdash\) \(f (a)\)
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\(=\) { \(A \subseteq A\) and definition of the inclusion map }
- \(f (\imath_{A \to A} (a))\)
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\(=\) { Definition 3.3.10 : function composition }
- \((f \circ \imath_{A \to A}) (a)\)
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\(=\) { \(f (a) \in B\) and definition of the inclusion map }
- \((\imath_{B \to B}) (f (a))\)
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\(=\) { Definition 3.3.10 : function composition }
- \((\imath_{B \to B} \circ f) (a)\)
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\(\vdash\) { Definition 3.3.7 (Equality of functions) }
- \(f = f \circ \imath_{A \to A} = \imath_{B \to B} \circ f\)
- \(\bullet\) (c) Show that, if \(f : A \to B\) is a bijective function, then \(f \circ f^{-1} = \imath_{B \to B}\) and \(f^{-1} \circ f = \imath_{A \to A}\).
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\(\Vdash\) { Prove each case }
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\(\bullet\) Show that \(f \circ f^{-1} = \imath_{B \to B}\)
- \(-\) Let \(b\) be any element in \(B\)
- ⊩ \((f \circ f^{-1})(b)\)
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= { Exercise-3.3.6 }
- \(b\)
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= { Definition of the inclusion map }
- \(\imath_{B \to B}(b)\)
- \(\square\)
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\(\bullet\) Show that \(f^{-1} \circ f = \imath_{A \to A}\)
- \(-\) Let \(a\) be any element in \(A\)
- ⊩ \((f^{-1} \circ f)(a)\)
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= { Exercise-3.3.6 }
- \(a\)
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= { Definition of the inclusion map }
- \(\imath_{A \to A}(b)\)
- \(\square\)
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\(\bullet\) Show that \(f \circ f^{-1} = \imath_{B \to B}\)
- \(\square\)
- \(\bullet\) (d) Show that there is a unique function \(h : X \cup Y \to Z\) such that \(h \circ \imath_{X→X \cup Y} = f\) and \(h \circ \imath_{Y \to X \cup Y} = g\) if
- \(-\) \(X\) and \(Y\) are disjoint sets
- \(-\) \(f : X \to Z\) and \(g : Y → Z\)