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**sapsapz** Let f: A->B be onto. We need to prove that exists g: B->A such that for all b in B, f o g (b) = b.

Let b1 be an arbitrary element of B.

f is onto, so there exists (at least one element called) a1 of A such that f(a1)=b1. Lets define g such that g(b1)=a1.

so f o g (b1) = f(g(b1))=f(a1)=b1.

b1 was arbitrary, and so it is true for all b in B. QED.

Dont I encounter a problem because two a elements might go to the same b? Or is it still ok because b is an arbitrary element?