0

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.

ffry
  • 15
  • Is there a question? Assuming there is, what have you tried? – Eric Towers Jun 15 '14 at 22:55
  • 1
    You should avoid using "inverse function theorem", because it already means something very different from this. I suggest renaming this as "Question on inverse functions" or something like that. – Luiz Cordeiro Jun 15 '14 at 22:57
  • I suppose you're talking about composition and inverses of relations (and, in particular, of functions), right? If not, would you care to give definitions? – Luiz Cordeiro Jun 15 '14 at 23:19
  • yes inverses of functions and believe it uses property of relations. I know it needs to be one-to-one and onto with g∘f=iA and f∘g=iB, which means g=f−1. The previous problem we were to apply that 'theorem' but now we need to find an alternative to it, and I have no idea... – ffry Jun 15 '14 at 23:50

1 Answers1

0

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 $V$, $$U\subset V\;\text{and}\;V\subset U\implies U=V$$