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

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  • $\begingroup$ You can prove this directly using nothing more than definitions. $\endgroup$ – Randall Jun 1 at 13:15
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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: $$h(a):=b_{g(a)}$$

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  • $\begingroup$ 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. $\endgroup$ – Mark Kamsma Jun 1 at 14:52
  • $\begingroup$ @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. $\endgroup$ – drhab Jun 1 at 14:57
  • $\begingroup$ I will post it as an answer in a bit then :) $\endgroup$ – Mark Kamsma Jun 1 at 15:19
  • $\begingroup$ @MarkKamsma Btw, proving that epis are split in $\mathbf{Set}$ also requires the axiom of choice. $\endgroup$ – drhab Jun 1 at 15:20
  • $\begingroup$ True, that was kind of my point. If you use it anyway, might as well use it that way. $\endgroup$ – Mark Kamsma Jun 1 at 15:54
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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 $\mathbf{Set}$.

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