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If $f \circ g$ is invertible, is $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$?

If not can someone give me a counterexample?

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    $\begingroup$ A formula which "works" on the real line is $f=\frac 1 x$, but this is not a function as it is undefined at $x=0$. The question does not specify what kinds of objects $f \text{ and } g $ are supposed to be. Precision of language would be easier to specify if the question were more precisely stated. $\endgroup$ – Mark Bennet Jun 25 '12 at 7:08
  • $\begingroup$ Also see math.stackexchange.com/q/2349/8581 $\endgroup$ – mrs Jun 25 '12 at 7:13
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It is true if $f$ and $g$ are invertible. You can check this directly by using the definition of inverse function.

But neither $f$ nor $g$ need to be invertible for $f\circ g$ to be invertible. To find a counterexample, I suggest looking for a noninjective $f$ and a nonsurjective $g$.

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  • $\begingroup$ I want to add this point to Jonas answer that we should have: $f:A\longrightarrow B$ and $g:B\longrightarrow C$ wherein $A,B$ and $C$ are sets. :) $\endgroup$ – mrs Jun 25 '12 at 7:06
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Let $ f:\{0,1\} \rightarrow \{0\} $ be defined by $ f(0) = 0 = f(1) $

and define $ g:\{0\} \rightarrow \{0\} $ with obvious mapping $g(0) = 0$

then $ f \circ g $ is defined from $ \{0\} $ to $ \{0\} $ with $ (f \circ g) (0) = 0 $ and $ (f \circ g)^{-1} (0) = 0 $

but f is not invertible.

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In my view. I think if we take them as binary relations, then we would have

$g^{-1} \circ f^{-1}$

=$\left \{(z,x) \in \operatorname{dom}(f^{-1}) \times \operatorname{ran}(g^{-1})|\exists y \in \operatorname{ran}(f^{-1}) \cap \operatorname{dom}(g^{-1})((z,y) \in f^{-1}\land(y,x) \in g^{-1})\right \}$

=$\left \{(z,x) \in \operatorname{ran}(f) \times \operatorname{dom}(g)|\exists y \in \operatorname{dom}(f) \cap \operatorname{ran}(g)((y,z) \in f\land(x,y) \in g)\right \}$

=$\left (\left \{(x,z) \in \operatorname{dom}(g) \times \operatorname{ran}(f)|\exists y \in \operatorname{dom}(f) \cap \operatorname{ran}(g)((y,z) \in f\land(x,y) \in g)\right \}\right )^{-1}$

=$(f\circ g)^{-1}$

That means $g^{-1} \circ f^{-1}=(f\circ g)^{-1}$ always holds regardless of whether they are invertible functions


For the direct proof of invertible function. We need to prove

  1. $dom((f \circ g)^{-1})=dom(g^{-1} \circ f^{-1})$

  2. for all $z \in dom((f \circ g)^{-1})$, $(f \circ g)^{-1}(z)=g^{-1} \circ f^{-1}(z)$

proof.

(1) Suppose that $z \in dom((f \circ g)^{-1})$, let $x=(f \circ g)^{-1}(z)$. Then $f \circ g(x)=z$. Hence $f^{-1}(z)=g(x)$. Consequently $g^{-1} \circ f^{-1}(z)=x$. Therefore $z \in dom(g^{-1} \circ f^{-1})$.

Suppose that $z \in dom(g^{-1} \circ f^{-1})$, let $x=g^{-1} \circ f^{-1}(z)$, analogously $x=(f \circ g)^{-1}(z)$.

Therefore $dom((f \circ g)^{-1})=dom(g^{-1} \circ f^{-1})$.

(2) According to the proof of (1) $g^{-1} \circ f^{-1}(z)=(f \circ g)^{-1}(z)$ for all $z$ in the domain of $(f \circ g)^{-1}$.

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