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THEOREM: if $F$ and $G$ are one to one then $G \circ F$ is also one to one and $(G \circ F)^\neg$ = $F^\neg \circ G^\neg$

PROOF:
if $F: A\rightarrow B$, $G: B \rightarrow C$ and

$\forall a, a' \in A \ \ F(a)=F(a') \Rightarrow a=a'$ then F is one to one and if

$\forall b, b' \in B \ \ G(b)=G(b') \Rightarrow b=b'$ then G is one to one by the definition of one to one.

If $(G \circ F)(a)=(G \circ F)(a') \Rightarrow G(F(a))= G(F(a'))$ then $F(a)= F(a')$ since $G$ is one to one.

If $F(a)=F(a')$ then $a=a'$ since $F$ is one to one

Because $G \circ F$ is one to one it is also invertible so $(G \circ F)^\neg$ exist now if we compute $\begin{align}((G\circ F)\circ (F^\neg\circ G^\neg))(a)&=\\ ((G\circ (F\circ F^\neg)\circ G^\neg))(a)&=\\ ((G\circ G^\neg))(a)&=a\end{align} $

So, $(F^\neg\circ G^\neg)$ is the inverse of $G \circ F$ therefore $(G \circ F)^\neg=(F^\neg\circ G^\neg)$

QED

I feel like the first part is ok but the 2nd part is all messed up ... what did I do wrong?

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    i couldn't figure out how to write f^{-1}2012-09-11

1 Answers 1

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I don't see anything wrong, though you probably ought to show that $((F^\neg\circ G^\neg)\circ(G\circ F))(a)=a,$ or at least argue that the demonstration is similar to going the other direction.

It's also worth noting (as others have pointed out in the comments) that we're really talking about $(G\circ F)^\neg:D\to A$, where $D:=G(F(A)).$ It seems like your source (or perhaps just you) may not be overly concerned with domains/codomains of functions, and is (are) simply leaving them as understood. One should be cautious about such things, though (as evidenced by the comments above).