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.


2 Answers 2


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. $\endgroup$
    – qualcuno
    Commented Jan 18, 2019 at 17:34
  • 1
    $\begingroup$ 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
    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}


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