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Proposition: For every A, B, f: A -> B, we have
1A: A -> A, 1A(x) := x
If f is injective, then we have
f o 1B = f 1A o f = f.
If f is also surjective (i.e., bijective), then we have
f o f-1 = 1A
f-1 o f = 1B.