I have no idea of how to do this. We need to create an alternative proof using some of the ideas on the bottom, but I'm lost. Any ideas on how to do this? I'm not sure how to even start the problem...
We are asked to use the following strategy to give an alternative proof of the following Theorem:
Suppose $f: A\rightarrow B, g:B \rightarrow A, g\circ f=i_A$ and $f\circ g=i_B.$ Then, $g=f^{-1}.$
Now we need to create an alternative to the above, using the information below.
Let $(b,a)$ be an arbitrary element of $B\times A.$ Assume $(b,a)\in g$ and prove $(b,a)\in f^{-1}.$ Then assume $(b,a)\in f^{-1}$ and prove $(b,a)\in g.$
Any approaches, information, outline of step, or ANYTHING you can give me will certainly help. No ideas on how to do this.
Assume that $(b,a)\in g$. This means, by definition, that $g(b)=a$. Since $f\circ g=i_B$, we also have that $f(g(b))=b$. Hence $f(a)=b$, so (again by definition) $(a,b)\in f$. And if $(a,b)\in f$, then $(b,a)\in f^{-1}$.
The other case will be similar. Now use the fact that for two sets $U$ and $V$, $$U\subset V\;\text{and}\;V\subset U\implies U=V$$