Category Theory – Finite Sets as Compact Objects in the Category $\mathrm {\mathbf{Sets}}$

category-theorylimits-colimits

Object $C$ of category $\mathcal C$ is called compact if
$$\mathrm{Hom} (X, \mathrm{colim} _I Y_i) \cong \mathrm{colim} _I \mathrm{Hom} (X, Y_i) $$
for every filtered colimit.

I want to prove that finite sets are compact in the category $\mathrm {\mathbf{Sets}}$ (the converse is also true as I know). Here is my attempt:

Every set $X=\{ x_j \}_{j\in J}$ can be described as a disjoint union $X\cong \sqcup _{j\in J} \{ * \}_j$. For finite sets this union is finite and hence commutes with filtered colimits.

For every set it is also true that $\mathrm{Hom} (\{ *\}, X) \cong X$.

Hence for every finite set $C=\{c_j\}_{j\in J}, |J|< \inf$

$$\mathrm{Hom} (C, \mathrm{colim}_i, Y_i)\cong \mathrm{Hom} (\sqcup_{j}\{ *\} , \mathrm{colim}_i, Y_i)\cong \prod_j \mathrm{Hom} (\{ *\}_j , \mathrm{colim}_i, Y_i) \cong \prod_j ( \mathrm{colim} _i Y_i)_j \overset{F(i, j)=Y_i}{\cong} \prod_j \mathrm{colim} _i F(i, j)\overset{\text{fin.lims comm. w. filt.colims}}{\cong}\mathrm{colim}_i \prod_j F(i, j)\cong \mathrm{colim} _i \mathrm{Hom} (\{ *\}_j, \prod F(i, j))\overset{\text{is it OK?}}{\cong}\mathrm{colim} _i \mathrm{Hom} (\sqcup_j \{ *\}_j, \prod F(i, j))\cong \mathrm{colim} _i \mathrm{Hom} (C, Y_i)$$

Upd: I think that is way better
$$\mathrm{Hom} (C, \mathrm{colim}_i, Y_i)\cong \mathrm{Hom} (\sqcup_{j}\{ *\} , \mathrm{colim}_i, Y_i)\cong \prod_j \mathrm{Hom} (\{ *\}_j , \mathrm{colim}_i, Y_i) \cong \prod_j ( \mathrm{colim} _i Y_i)_j\cong\prod_j \mathrm{colim} _i \mathrm{Hom} (\{ *\}_j, Y_i) \overset{\text{fin.lims comm. w. filt.colims} }{\cong} \mathrm{colim}_i \prod_j \mathrm{Hom} (\{ *\}_j, Y_i)\cong \mathrm{colim}_i \mathrm{Hom} (\sqcup_j \{ *\}_j, Y_i)\cong\mathrm{colim}_i \mathrm{Hom} (C, Y_i). $$

Upd2: There is an obvious typo after "is it OK?": there should be no product, only coproduct. But it still doesn't work that way.

Best Answer

The idea of the proof is OK, but the execution is not (exactly where you write "is it OK").

But you can write down the proof in a more conceptual way. Prove the following:

  1. In any category, compact objects are closed under finite colimits. This essentially follows from the fact (*) you already cited: finite limits commute with filtered colimits in $\mathbf{Set}$.

  2. The terminal set $1$ is compact in $\mathbf{Set}$.

  3. Every finite set is a finite coproduct of copies of $1$.

The result follows immediately. Notice however that a direct proof is also perfectly possible using the explicit construction of the filtered colimit of sets, and I believe that the same argument is happening in the fact (*) you are already using,

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