Jections
Properties of Surjections, Injections and Bijections
Let z an arbitrary element in C.
Then since f is a surjection, there is an element y in B such that z = f(y).
Then since g is a surjection, there is an element x in A such that y = g(x).
Hence by the definition of composite function, z = f(g(x)), that is z = fg(x).
Hence fg is a surjection.
QED
Let x1 and x2 be arbitrary elements in A.
Suppose that f(g(x1)) = f(g(x2)).
Then since f is an injection, g(x1) = g(x2).
Then since g is an injection, x1 = x2.
That is for any pair of elements x1 and x2 in A if f(g(x1)) = f(g(x2)), then x1 = x2.
Hence fg is an injection.
QED
Let z an arbitrary element in C.
Then since fg is a surjection, there is an element x in A such that z = f(g(x)).
By the definition of fg, there is an element y in B such that y = g(x) and z = f(y).
Hence for an arbitrary element z in C, there is an element y in B such that z = f(y).
Hence f is a surjection.
QED
Proof is left as an exercise.
Since f( S ∩ T ) ⊆ f(S) ∩ f(T) for a function f, we need to prove that f(S) ∩ f(T) ⊆ f( S ∩ T ) for a bijection f.
Let y be an arbitrary element of f(S) ∩ f(T).
Then there is an element x1 in S and an element x2 in T such that y = f(x1) = f(x2).
Since f is a bijection, it is an injection.
Hence if f(x1) = f(x2), then x1 = x2.
Hence x1 (= x2) ∈ S ∩ T.
Hence y (= f(x1) = f(x2)) ∈ f( S ∩ T ).
Hence f(S) ∩ f(T) ⊆ f( S ∩ T ) if f is a bijection.
QED