Given any real function, if $f(f(x))=x$ does that mean $f(x)$ is its own inverse? I am confused since $f^{-1}(f(x))=x$ and this is a fact, so can we assume that $f(x)$ will equal $f^{-1}(x)$ by substitution? Specify if possible if it is never, sometimes, or always true.
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@user148177 But $f(x)=|x|$ does not satisfy $f(f(x))=x$. – Michael Burr Jun 30 '15 at 18:41
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@user148177: that doesn't work. $f(f(-1))=1 \neq -1$ – Ross Millikan Jun 30 '15 at 18:41
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When you write $f-1(f(x))$ do you mean the function $(f-1)(x):=f(x)-x$? – Michael Burr Jun 30 '15 at 18:43
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The first sentence is true. However, it's not always true that a function has an inverse to begin with. – dannum Jun 30 '15 at 18:48
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The statement is true, but it is not valid to use substitution when dealing with functions. – Michael Burr Jun 30 '15 at 18:49
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Let $f:A\rightarrow A$ with $f\circ f=1_A$ (The identity in $A$)
If $f$ is not surjective then $f\circ f$ is not surjective and if $f$ is not injective $f\circ f$ is not injective.
Hence if $f\circ f$ is bijective then $f$ is bijective. $f$ must therefore have an inverse $f^{-1}$. Using this we obtain the following:
$f\circ f=1_a\implies f^{-1}\circ f\circ f=f^{-1}\circ 1_A\implies 1_A\circ f=f^{-1}\implies f= f^{-1}$
Asinomás
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4It's simpler than that: the inverse, by definition, is the function satisfying $$ f \circ f^{-1} = f^{-1} \circ f = 1 $$ $f$ fits the requirement for $f^{-1}$ – Ben Grossmann Jun 30 '15 at 18:54
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Oh, good point. you should add that as a solution, then the only thing that might be worth proving is that inverses are unique when they exist. – Asinomás Jun 30 '15 at 18:59