I'm currently working on the following theorem from Emily Riehl's Category Theory in Context:

Theorem 3.8.9. Filtered colimits commute with finite limits in $\mathbf{Set}$.

I understand most of the proof, except for a small detail which I will proceed to explain next. The proof starts by considering the canonical map $$ \kappa : \text{colim}_J \text{lim}_IF(i,j) \longrightarrow \text{lim}_I\text{colim}_JF(i,j) \tag{1} $$ and trying to prove that it is a bijection. The author also previosuly highlights a key remark, which can be paraphrased as:

When $J$ is a small filtered category, for any functor $G \in \mathbf{Set}^J$ we have that $\text{colim}_JG = \big(\coprod_j Gj\big) / \sim$ with $x \in Gj \sim y \in Gk$ if and only if we have arrows $f : j \to t, g : k \to t$ such that $Gf(x) = Gg(y)$.

To see that $\kappa$ is surjective, we consider an element of $\text{lim}_I\text{colim}_JF(i,j)$. Since we are working with the limit of a set valued funtor, this is equivalent to choosing a cone $(\lambda : 1 \Rightarrow \text{colim}_JF(i,-))_{i\in I}$. Each $\lambda_i$ corresponds by the remark to an element of $\text{colim}_JF(i,-)$ which can be identified as the class of equivalence of a certain element $\lambda_i \in F(i,j_i)$. Now, the author claims that we can find $t \in J$ so that each $\lambda_i$ is equivalent to some $\lambda'_i \in F(i,t)$ and that moreover, $(\lambda': 1 \Rightarrow F(-,t))_{i \in I}$ is a cone.

The first part of this claim, I do have managed to prove: since $I$ is finite, the full subcategory $J_I$ spanned by the $(j_i)_i$ can be thought as a diagram $I \to J$. The latter is filtered and $I$ is finite, so it follows that we have a cone $(\mu_i: j_i \to t)_{i \in I}$ under the diagram. Defining $\lambda'_i$ to be $F(1_i,\mu_i)(\lambda_i)$, we see that $\lambda_i \sim \lambda'_i$ via the arrows $1_{t}$ and $\mu_i$.

From here on, I have unable to finish the proof of the claim, that is, I haven't been able to prove that $\lambda'$ is a cone. I have tried to write maps $\lambda'_i : 1 \to F(i,t) $ as compositions $F(1_i,\mu_i)\lambda_i$ and then using that $\lambda$ is itself a cone (on colimit objects), but I have not been able to deal with the fact that these factorizations change the object in $J$ (for example, to $j_i$ and $j_{i'}$, if we deal with an arrow $g : i\to i'$ to begin with).

Any ideas on how to conclude from here? Thanks in advance.


For any functor of the form $U:\mathcal A\times\mathcal B\to\mathcal C$ and any morhpisms $\alpha:a\to a',\ \beta:b\to b'$, we have the following symmetry: $$U(\alpha,b')\,U(a,\beta)\ =\ U(\alpha,\beta)\ =\ U(a',\beta)\,U(\alpha,b)\ ,$$ where e.g. $U(\alpha,b)$ stands for $U(\alpha,1_b)$

simply because we already have $(\alpha,1_{b'})(1_a,\beta)=(\alpha,\beta)=(1_{a'},\beta)(\alpha,1_b)$ in $\mathcal A\times\mathcal B$.

Now, let $\gamma:i\to i'$ be an arrow in $I$, then we have $$F(\gamma,t)\,\lambda_i'\ =\ F(\gamma,t)\,F(i,\mu_i)\,\lambda_i\ =\ F(\gamma,\mu_i)\,\lambda_i\ =\\ =\ F(i',\mu_i)\,F(\gamma,j_i)\,\lambda_i\ \overset{\lambda\text{ cone}}=\ F(i',\mu_i)\,\lambda_{i'}\ =\ \lambda'_{i'} $$

  • $\begingroup$ Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it. $\endgroup$ – Guido A. Jan 18 at 17:34
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    $\begingroup$ The part with the $I\to J$ diagram and especially with $J_I$ was indeed unclear until I wrote that part with my words in the answer. Then I realized it was just the same argument and deleted it.. $\endgroup$ – Berci Jan 18 at 18:30

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