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$f: X \to Y$ and $g: Y \to Z$ are injective functions.(for all $x_1x_2∈X$ if $f(x_1)=f(x_2)$ then $x_1=x_2$) How can I prove, that $ (g \circ f)^ {-1}=f^ { -1 }\circ g^{-1 }$? (where by $f^ { -1 }$ I mean the inverse of $f$ http://en.wikipedia.org/wiki/Inverse_function )

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Well, $h^{-1}(b)=a \ \iff\ b=h(a)$ for an injective $h$ (so that $a$ is unique for any $b$ in the range of $h$).

So, we get $$(g\circ f)^{-1}(z)=x\ \iff \ z=\left(g\circ f\right)(x) \ \iff\ z=g(f(x))\ \iff \dots$$

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  • $\begingroup$ I appreciate your help, but I don't know how I should continue and finish your proof. I am assuming the next step following your logic would be $ x=g(f(z))^{-1}$? But I don't know how I should continue... I am sorry, I am aware I look really dumb now, but I am (very) new to mathematics. Thanks again for your help. $\endgroup$ – user176791 Sep 23 '14 at 19:20
  • $\begingroup$ Next equation to write is $g^{-1}(z)=f(x)$. $\endgroup$ – Berci Sep 23 '14 at 21:30
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Hint:

If $x \in Z\setminus R((g \circ f)$ then it will be $\phi$.

If $x \in R((g \circ f)$ then check what will be image of $ (g \circ f)^ {-1}(x)$ & $f^ { -1 }\circ g^{-1 }(x)$

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