# About a specific step in a proof of the fact that filtered colimits and finite limits commute in $\mathbf{Set}$

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 $$F(\gamma,t)\,\lambda_i'\ =\ F(\gamma,t)\,F(i,\mu_i)\,\lambda_i\ =\ F(\gamma,\mu_i)\,\lambda_i\ =\\ =\ F(i',\mu_i)\,F(\gamma,j_i)\,\lambda_i\ \overset{\lambda\text{ cone}}=\ F(i',\mu_i)\,\lambda_{i'}\ =\ \lambda'_{i'}$$

• Neat! I got it now. Thanks for taking the time to untangle my thoughts, I appreciate it. Commented Jan 18, 2019 at 17:34
• 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.. Commented Jan 18, 2019 at 18:30

I have the same question as guidoar, but I don't quite follow Berci's answer. In particular, the last equation: $$F(i',\mu_i)\lambda_{i'} = \lambda'_{i'}$$ seems not to make sense, as $$\lambda_{i'}$$ is in $$F(i',j_{i'})$$, but the domain of $$F(i',\mu_i)$$ is $$F(i',j_i)$$. What we need to show is rather: $$F(\gamma, \mu_i)\lambda_i = F(i',\mu_{i'})\lambda_{i'}$$

I think the trick is to consider not just the (discrete) cocone under the $$j_i$$'s but a larger cocone with data about the morphisms of $$I$$.

Because the $$(i,\lambda_i)$$'s form a cone, it follows that for any morphism $$\gamma : i \to i'$$ there exists some object $$j_\gamma \in J$$ and morphisms: $$\sigma_{\gamma} : j_i \to j_{\gamma} \quad \tau_{\gamma} : j_{i'} \to j_f$$ such that: $$F(\gamma,\sigma_{\gamma})\lambda_i = F(i', \tau_{\gamma})\lambda_{i'}$$

Since $$I$$ has finitely many morphisms, we can construct the cocone $$\mu$$ under the diagram of all the $$j_i$$'s and $$j_{\gamma}$$'s with the $$\sigma_\gamma$$'s and $$\tau_{\gamma}$$'s. In particular, we have for this cocone that for any $$\gamma : i \to i'$$, $$\mu_i = \mu_{\gamma} \circ \sigma_{\gamma}$$ and $$\mu_{i'} = \mu_{\gamma} \circ \tau_{\gamma}$$

Now we can show that: $$F(\gamma, \mu_i)\lambda_i = F(i',\mu_{i'})\lambda_{i'}$$ as follows: \begin{align} F(\gamma, \mu_i)\lambda_i = & F(\gamma, \mu_\gamma \circ \sigma_\gamma)\lambda_i \\ = & F(i',\mu_\gamma)(F(\gamma,\sigma_\gamma)\lambda_i) \\ = & F(i',\mu_\gamma)(F(i',\tau_{\gamma})\lambda_{i'}) \\ = & F(i',\mu_\gamma \circ \tau_\gamma)\lambda_{i'} \\ = & F(i',\mu_{i'})\lambda_{i'} \end{align}