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Let $X$ and $Y$ be arbitrary sets and $f:X\rightarrow Y$ an isomorphism. Prove that there exist a transformation $g:Y\rightarrow X$ such that $f\circ g$ is the identity in $Y$.

As X and Y havn't a structure to preserve, $f$ is just a bijective function; hence exist a function $g$ such that $f\circ g$ is the identity in Y. In particular this function is a transformation.

José Osorio
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    You need to specify what you mean by "transformation" and what is you definition of isomorphism. – Caligula Jul 05 '16 at 08:03
  • My guess is that "transformation" and "isomorphism" should both be replaced by "function". – Did Jul 05 '16 at 08:21

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