# Covariant representable functors from $Set$ to $Set$ preserve surjective maps. [closed]

Show that covariant representable functors from $$Set$$ to $$Set$$ preserve surjective maps. Give me a hint please.

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• You can prove this directly using nothing more than definitions. – Randall Jun 1 at 13:15

Let $$A,B,C$$ be sets and let $$f$$ be a surjective function $$B\to C$$.

Then it is enough to prove that for every function $$g:A\to C$$ there is a function $$h:A\to B$$ such that: $$g=f\circ h$$

This means that $$\mathsf{Hom}(A,f):\mathsf{Hom}(A,B)\to\mathsf{Hom}(A,C)$$ is surjective whenever $$f$$ is surjective.

Try to find $$h$$ yourself (you asked for a hint).

If you don't manage then have a look at the rest.

For every $$c\in C$$ choose (AC is used here) an element $$b_c\in B$$ that satisfies $$f(b_c)=c$$ and prescribe $$h$$ by stating that: $$h(a):=b_{g(a)}$$

• You do use the axiom of choice here (which may be worth noting). In that case you can also use the fact that every epi in $\mathbf{Set}$ splits, and that functors preserve split epis. Which is a special case of your proof, but perhaps a bit easier. – Mark Kamsma Jun 1 at 14:52
• @MarkKamsma Yes, the axiom of choice is used. What you propose is indeed more elegant and more general (the functor does not have to be representable). It only requires a proof that in category $\mathbf{Set}$ every epi is a retraction, which is not difficult. You could give that as an alternative answer yourself. I will certainly upvote it. – drhab Jun 1 at 14:57
• I will post it as an answer in a bit then :) – Mark Kamsma Jun 1 at 15:19
• @MarkKamsma Btw, proving that epis are split in $\mathbf{Set}$ also requires the axiom of choice. – drhab Jun 1 at 15:20
• True, that was kind of my point. If you use it anyway, might as well use it that way. – Mark Kamsma Jun 1 at 15:54

There is already a good answer by drhab, but as discussed in the comments there might be an easier proof.

We call an arrow $$f: A \to B$$ a split epimorphism if there is an arrow $$s: B \to A$$ such that $$fs = Id_B$$. One can easily show that every split epimorphism is indeed an epimorphism (exercise, hover over the yellow box for a proof).

Let $$f: A \to B$$ and $$s: B \to A$$ be such that $$fs = Id_B$$. Let now $$g,h: B \to C$$ be such that $$hf = gf$$. Then $$h = hfs = gfs = g$$, so $$f$$ is indeed an epimorphism.

Then, by the axiom of choice, every epimorphism (i.e. surjective function) in $$\mathbf{Set}$$ splits. Functors always preserve split epimorphisms (again, exercise, hover over the yellow box for a proof).

Suppose we have a functor $$F: \mathcal{C} \to \mathcal{D}$$, and a split epimorphism $$f: A \to B$$ in $$\mathcal{C}$$. Let $$s: B \to A$$ be such that $$fs = Id_B$$. Then for $$F(f): F(A) \to F(B)$$ we have that $$F(s): F(B) \to F(A)$$ is such that $$F(f)F(s) = F(fs) = F(Id_B) = Id_{F(B)}$$, so $$F(f)$$ is indeed a split epimorphism.

So in particular, any functor $$F: \mathbf{Set} \to \mathbf{Set}$$ preserves split epimorphisms, which coincide with surjective functions in $$\mathbf{Set}$$.