“…Note that (Φ( 1 )( ), … , Φ( )( )) = (Φ( 1 )( ), … , Φ( )( )), and this implies that the compounds are equal, so Φ( )( ) = Φ( )( ) for each , but the map Φ is injective so ( ) = ( ) for each ∈ {1, … , } this shows that = for each ∈ {1, … , } which implies that = which means that must be injective. Using the results from 10 , it can be seen that is invertible, and this completes our proof.…”