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Not necessarily true. Let A = {1,2,3}, B = {4,5,6}, C = {7,8,9}. Lets define the following functions:

f(1) = 4, f(2) = 4, f(3) = 4, that is, f maps all elements in A to 4 in B.

Then define g(4) = 7, g(5) = 8, g(6) = 8.

Also define h(4) = 7, h(5) = 9, h(6) = 9. Then clearly h not= g, as g = {(4,7), (5,8), (6,8)}, h = {(4,7), (5,9), (6,9)}.

Then we see that g ∘ f= h ∘ f = 7 for all a in A, however g not= h, so this is a counterexample to the statement.

I think the statement is only true if f is onto B.

Proof: Let f be a surjective mapping from A onto B. Then for all b in B, b = f(a) for some a in A.

Now using point-wise evaluation, choose any b in B.

Then g(b) = g(f(a)) for some a in A, and g(f(a)) =(g ∘ f)(a)= (h ∘ f)(a)=h(f(a)) = h(b). Since our choice of b was arbitrary, g(b) = h(b) for all b in B, hence g = h.

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