# If $F$ and $G$ are one to one, then $G \circ F$ is one to one and  $(G \circ F)^\neg = F^\neg \circ G^\neg$

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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In order for the inverse function to exist you need bijectivity, also called "one-to-one and onto". You must prove that every point in the target space is in the image of $G\circ F$. (The statement of the theorem suggests that that's what you mean by "one to one" here, but you seem to prove injectivity only.) – Sean Eberhard Sep 10 '12 at 19:59
Please try to use more descriptive titles. – Austin Mohr Sep 10 '12 at 20:02

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The proof of one to one has stylistic deficiencies. For one thing, the first five lines are unnecessary, and somewhat confusing. Also, symbols from logic are not quite used correctly. For invertibility in the usual sense, you need more. Otherwise, best you can do is produce an inverse on a suitably restricted subset of $C$. – André Nicolas Sep 10 '12 at 20:02
Is the notation $f^\neg$ for inverse function common? I've always only seen $f^{-1}$. – Henning Makholm Sep 10 '12 at 20:24
i couldn't figure out how to write f^{-1} – Joshua Rocky Lizardi Sep 11 '12 at 2:07

## 1 Answer

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).

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