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Let $f: A\to B$ and that $f$ is a bijection. Show that the inverse of $f$ is bijective.

Surjectivity: Since $f^{-1} : B\to A$, I need to show that $\operatorname{range}(f^{-1})=A$. But since $f^{-1}$ is the inverse of $f$, and we know that $\operatorname{domain}(f)=\operatorname{range}(f^{-1})=A$, this proves that $f^{-1}$ is surjective.

Injectivity: I need to show that for all $a\in A$ there is at most one $b\in B$ with $f^{-1}(b)=a$. But we know that $f$ is a function, i.e. for all $a\in A$ there is exactly one (at least one and never more than one) $b\in B$ with $f(a)=b$. 'Exactly one $b\in B$' obviously complies with the condition 'at most one $b\in B$'.

Since $f^{-1}$ is the inverse of $f$, $f^{-1}(b)=a$. So combining the two, we get for all $a\in A$ there is exactly one (at least one and never more than one) $b\in B$ with $f^{-1}(b)=a$.

I think my surjective proof looks ok; but my injective proof does look rather dodgy - especially how I combined '$f^{-1}(b)=a$' with 'exactly one $b\in B$' to satisfy the surjectivity condition. Could someone verify if my proof is ok or not please? Thank you so much!

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    $\begingroup$ In stead of this I would recommend to prove the more structural statement: "$f:A\to B$ is a bijection if and only if it has an inverse". An inverse is a map $g:B\to A$ that satisfies $f\circ g=1_B$ and $g\circ f=1_A$. So it is immediate that the inverse of $f$ has an inverse too, hence is bijective. $\endgroup$ – drhab Nov 6 '15 at 9:55
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Your proof is logically correct (except you may want to say the "at least one and never more than one" comes from the surjectivity of $f$) but as you said it is dodgy, really you just needed two lines:

(1) $f^{-1}(x)=f^{-1}(y)\implies f(f^{-1}(x))=f(f^{-1}(y))\implies x=y$.

(2) Let $a\in A$, then $f^{-1}(f(a))=a$.

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  • $\begingroup$ I thought for injectivity it should be (in the case of the inverse function) whenever b≠b then f^-1(b)≠f^-1(b)? I am not sure why would f^-1(x)=f^-1(y)? Would you mind elaborating a bit on where does the first statement come from please? Also when you talk about my proof being logically correct, does that mean it is incorrect in some other respect? Thank you! $\endgroup$ – Daniel Mak Nov 6 '15 at 9:44
  • $\begingroup$ Do you know about the concept of contrapositive? $b\neq b \implies f^{-1}(b)\neq f^{-1}(b)$ is logically equivalent to $f^{-1}(b)= f^{-1}(b)\implies b=b$ $\endgroup$ – cr001 Nov 6 '15 at 9:50
  • $\begingroup$ Yes I know about that, but it seems different from (1). In the antecedent, instead of equating two elements from the same set (i.e. f^-1(b) and f^-1(b')), (1) is equating two different variables to each other (f^-1(x) and f^-1(y)), that's why I am not sure I understand where it is from. $\endgroup$ – Daniel Mak Nov 6 '15 at 12:48
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    $\begingroup$ There is never a need to prove $b\neq b \implies f^{-1}(b)\neq f^{-1}(b)$ because $b\neq b$ is never true in the first place. What we want to prove is $a\neq b \implies f^{-1}(a)\neq f^{-1}(b)$ for any $a,b$ $\endgroup$ – cr001 Nov 6 '15 at 13:03
  • $\begingroup$ Oooh I get it now! x and y are supposed to denote different elements belonging to B; once I got that outta the way I see how substituting the variables within the functions would yield a=a'⟹b=b', where a and a' belong to A and likewise b and b' belong to B. Thank you so much! This has been bugging me for ages so I really appreciate your help $\endgroup$ – Daniel Mak Nov 7 '15 at 10:01

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