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## Homework Statement

Suppose f: A → B is a function. Show that f is surjective if and only if there exists g: B→A such that fog=i

_{B}, where i is the identity function.

## The Attempt at a Solution

Well, I believe for a rigorous proof we need to use the axiom of choice, but because I have never worked with the axiom of choice before I need you guys to check if my proof is rigorous and correct.

1) Assume that f is onto.

Define the set ∏={f

^{-1}[{y}] : y ε B} (ε represents the membership relation). That means ∏ is the set of all pre-images of f restricted to the element y in B. It's obvious that ∏ is a family of sets and because its members are non-empty sets (since f is onto) we could apply the axiom of choice to it.

Axiom of choice tells us that there exists a function h: ∏ → U∏. It's obvious that U∏=A.

Now let's define: g:B → A , g(y)=h(f

^{-1}[{y}]). We should show that it's well-defined.

Which means if y=z then g(y)=g(z), but it's obvious, since if y=z then y and z are the same element in B, hence f

^{-1}[{y}]=f

^{-1}[{z}], since h is a function (AC), h(f

^{-1}[{y}])=h(f

^{-1}[{z}]) which means g(y)=g(z).

Now we need to prove that g is a right inverse for f.

Well, let's see what fog does to an arbitrary element y in B. by definition, g(y)=h(f

^{-1}[{y}]), then fog(y)=f(h(f

^{-1}[{y}])). Now let's see what's happening when fog acts on y. first it gives us the set of pre-image of y, then h chooses one element in f

^{-1}[{y}], then f acts on it and it's again sent to y! That's it. It proves that fog(y)=y and since y was arbitrary we conclude that fog=i

_{B}.

Now, suppose that there exists a function g such that fog(y)=y. We want to show that f is onto. For any y in B, we have: fog(y)=f(g(y))=y; now if we set g(y)=x we clearly see that for any y ε B, there exists x=g(y) such that f(x)=y which proves that f is onto.