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In the question is assumed that the function $f$ is a one-to-one function from $A$ onto $B$.

Here is my atempt:

By definition I have that: $f^{-1} = \{(x,y) : (y,x) \in f\}$ and from this I conclude that $(f^{-1})^{-1} = \{(x,y) : (y,x) \in f^{-1}\}$

First let some $t$ belong to $(f^{-1})^{-1}$ where $t=(a,b)$, from this we have that $(b,a) \in f^{-1}$, but if $(b,a)$ is in $f^{-1}$, then $(a,b) = t \in f$ and from this $(f^{-1})^{-1} \subseteq f.$

Now let some $t$ belong to $f$ with $t=(u,v)$, from this we have that $(v,u) \in f^{-1}$ and then $t = (u,v) \in (f^{-1})^{-1}$ and therefore $f \subseteq (f^{-1})^{-1}$ $$[((f^{-1})^{-1} \subseteq f) \land (f \subseteq (f^{-1})^{-1})] \Leftrightarrow (f^{-1})^{-1} = f$$

This question is under the exercises about composition of functions, should I have using some property of composition?

Paulo Henrique L. Amorim
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    This proof is ok! – Unknown Jul 19 '20 at 01:46
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    @AmanPandey You just need to state that for the function each $x$ has exactly one $y$. Else, you are proving the case for any relation. The inverse exists and is a function because $f$ is a bijection – M.O.Darwish Jul 19 '20 at 01:54
  • I got confuused by that similar question in his answers https://math.stackexchange.com/questions/209110/prove-that-the-inverse-of-an-inverse-is-the-function-itself-f-1-1-f ,because in the book Im following the past two question before this one was to prove that $F \circ F^{-1} = I_B$ and $F^{-1} \circ F = I_A$, where $I_X$ is the identity function in X – Paulo Henrique L. Amorim Jul 19 '20 at 01:56
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    @AmanPandey I missed to provide this in the question, sorry, this was placed as an Theorem in the book, where its proved that inverse of bijection will be a function and a bijection too. – Paulo Henrique L. Amorim Jul 19 '20 at 02:00
  • @PauloHenriqueL.Amorim If you realized. Then it's good. – Unknown Jul 20 '20 at 09:09

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