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Show that a map is invertible if and only if it is both injective and surjective.
Proof: \((\Longrightarrow)\) Assume that a map\[\begin{aligned} f:X&\to Y\\ x&\mapsto y \end{aligned} \]is invertible, then exists a map \(f^{-1}:Y\to X\), such that
\[f\circ f^{-1}=1_{Y},\quad f^{-1}\circ f=1_X. \]We need to show that \(f\) is both injective and surjective.
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Let \(x_1,x_2\in X\) and \(x_1\neq x_2\). Since \(f^{-1}\circ f=1_X\), we have
\[f^{-1}\circ f(x_1)=x_1\neq x_2=f^{-1}\circ f(x_2), \]i.e. \(f^{-1}(f(x_1))\neq f^{-1}(f(x_2))\). Thus \(f(x_1)\neq f(x_2)\), \(f\) is injective.
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Let \(y \in Y\), note that
\[y=f\circ f^{-1}(y)=f(f^{-1}(y)). \]There we find \(x:=f^{-1}(y)\in X\), such that \(f(x)=y\). Thus \(f\) is surjective. ■
\((\Longleftarrow)\) If \(f\) is both injective and surjective i.e. \(f\) is bijective, then \(f\) has a unique two-sided inverse \(f^{-1}\) from Corollary 1.2.20, i.e. \(f\) is invertible.
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Let \(X\) be a set, \(A\) be a subset of \(X\). Is \(f:P(X)\to P(X),~A\to A'\) a bijection? Why?
Solution: Obviously, \(f\) is a map since for each \(A\in P(X)\), there exists a unique \(A'\) corresponding to it through the correspondence rule. Therefore, we only need to check if \(f\) is both injective and surjective.- Let \(A, B\in P(X)\) and \(A\neq B\). Since \(A\neq B\), there must \(\exists ~x\in A\) and \(x\notin B\) or \(\exists ~y\in B\) and \(y\notin A\). Without loss of generality, suppose that \(\exists ~x\in A\) and \(x\notin B\), i.e. \(x\in A\text{\\}B\). Note that \(A\text{\\}B\subseteq X\text{\\}B=B'\), so \(x\in B'\). And it is also evident that \(x\notin A'\) since \(x\in A\). Thus \(A'\neq B'\), i.e. \(f\) is injective.
- Let \(A\in P(X)\), we have \(A'=X\text{\\}A\in P(X)\) and \(f(A')=(A')'=A\). Thus \(f\) is surjective.
In summary, \(f\) is bijective.
Reference: 王豔華 《抽象代數基礎》- Section 1.2