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Suppose that $f:A\rightarrow B$ and $g:B \rightarrow C$ are both one-to-one and onto. Prove that $gf$ is one-to-one and onto. Prove further that $(gf)^{-1} = f^{-1}g^{-1}$.

I have already proven the first part, but the second part has always puzzled me. I have tried assuming $x \in (gf)^{-1}$ but that doesn't lead to nowhere. Nor does $x \in (gf)^{-1}(t)$ and showing $x = t$. How to prove?

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marked as duplicate by Daniel Fischer Oct 3 '17 at 10:08

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The definition of the inverse is that $h \circ h^{-1} = \operatorname{id}$ and $h^{-1} \circ h = \operatorname{id}$, where $\operatorname{id}$ is the identity function.

Showing that $(gf)^{-1} = f^{-1}g^{-1}$ is equivalent to showing

$$g \circ f \circ f^{-1} \circ g^{-1} = \operatorname{id}$$ $$ f^{-1} \circ g^{-1} \circ g \circ f= \operatorname{id}$$

Can you do this?

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  • $\begingroup$ Do I have to define f as a 1-1 onto function first, and then same for g? @Isaac Solomon $\endgroup$ – Don Larynx Sep 28 '13 at 23:48
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    $\begingroup$ @DonLarynx $f$ and $g$ must be one-to-one and onto in order for the question to make sense, since $f^{-1}$ and $g^{-1]$ otherwise don't exist. $\endgroup$ – bradhd Sep 28 '13 at 23:56
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$\newcommand{\imp}{\quad\Longrightarrow\quad}$

$$ \left(\rm gf\right)^{-1}\left(y\right) = x \imp y = \left(\rm gf\right)\left(x\right) = {\rm g}\left({\rm f}\left(x\right)\right) \imp {\rm f}\left(x\right) = \left(\rm g\right)^{-1}\left(y\right) $$

$$ \imp x = \left(\rm f\right)^{-1}\left(\left(\rm g\right)^{-1}\left(y\right)\right) = \left(\left(\rm f\right)^{-1}\left(\rm g\right)^{-1}\right)\left(y\right) $$

$$\color{#ff0000}{\large% \left(\rm gf\right)^{-1}\left(y\right) = \left(\left(\rm f\right)^{-1}\left(\rm g\right)^{-1}\right)\left(y\right)} $$

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$(fg) x = f(g(x))$. $(fg)^{-1} (fg) x = (fg)^{-1} f(g(x))$. If you want to get back to $x$, what function do you first have to apply to $f(g(x))$? What function do you have to apply next?

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