(Q7): ∀ y ∈ T ( f(g(y)) = y ) ∧ ∀ x ∈ S ∃ y ∈ T ( g(y) = x ) ⇒ ∀ x,y ∈ S (f(x) = f(y) ⇒ x = y), where f : S -> T and g : T -> S
If ∀ y ∈ T ( f(g(y)) = y ) ∧ ∀ x ∈ S ∃ y ∈ T ( g(y) = x ):
	∀ y ∈ T ( f(g(y)) = y )
	∀ x ∈ S ∃ y ∈ T ( g(y) = x )
	Given a ∈ S:
		Given b ∈ S:
			If f(a) = f(b):
				∃ y ∈ T ( g(y) = a )
				∃ y ∈ T ( g(y) = b )
				Let a' ∈ T such that g(a') = a
				Let b' ∈ T such that g(b') = b
				g(a') = a
				g(b') = b
				f(g(a')) = a'
				f(g(b')) = b'
				f(a) = a'
				f(b) = b'