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I've read that the composition of functions is not commutative but in this case it works. So how can we proof that the composition of a function and its inverse commutative?

GniruT
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  • Because the composition of a function and its inverse is the identify function, regardless of what order you do it in. Though those identify functions are only equal if the domain and range of the functions is the same. – Gregory Grant Nov 09 '15 at 23:50
  • But is it not necessary a proof? Then if the domain and range of the functions is the same how can we proofe that the identity functions are the same? – GniruT Nov 09 '15 at 23:54
  • By definition: the inverse function $g$ of the function $f$ is the unique one such that $g\circ f$ and $f\circ g$ are identity functions. – egreg Nov 10 '15 at 00:03
  • On the other hand, you can prove that if a function admits a left and a right inverse, then they must be the same. – A.P. Nov 10 '15 at 00:30

1 Answers1

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By definition a function $g:X\to Y$ is the inverse of a function $f:Y\to X$ if $f\circ g$ is the identity map of $X$ and $g\circ f$ is the identity map of $Y$.

  • If $X$ and $Y$ are not the same set, then it makes no sense to say that $f$ and $g$ are commutative, for the result of the two compositions $f\circ g$ and $g\circ f$ cannot be equal simply because their domains and codomains are different.

  • If, on the other hand, $X$ and $Y$ are the same set, then $f$ and $g$ are commutative simply by definition!

Mariano Suárez-Álvarez
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  • Thanks! I've forgot the "meaning" of the inverse. Now it's clear. Is the inverse bijektive as well? Should we proof it or is it so by definition? – GniruT Nov 10 '15 at 02:12