I have a question involving preservation of cofiltered limits. Ordinarily this would be a very boring question, but it comes up in condensed math in its analogue of the completeness concept.
The concept is that of "solid", which says that for each profinite set $S = \text{lim} S_i$, maps of condensed abelian groups $\mathbb{Z}[S] \rightarrow M$ lift against the canonical map $\mathbb{Z}[S] \rightarrow \text{lim} \mathbb{Z}[S_i]$.
One broad goal of condensed math is to extend commutative algebra to subsume some areas of functional analysis. The concept of solid plays a role similar to completion. But I don't as of yet understand the condition very well.
I don't understand why the free condensed abelian group $\mathbb{Z}[S]$ is not identically $\text{lim} \mathbb{Z}[S_i]$. I am trying to come up with a broad class of examples as follows:
Let $\mathcal{S}$ be a site. Consider the category of sheaves of sets on the site $\mathcal{S}$, $\text{Sh}(\mathcal{S}, \text{Set})$, and the category of sheaves of abelian groups $\text{Sh}(\mathcal{S}, \text{Ab})$ on the site $\mathcal{S}$. I am looking for a general class of examples showing that the free functor $\text{Sh}(\mathcal{S}, \text{Set}) \rightarrow \text{Sh}(\mathcal{S}, \text{Ab})$ does not preserve cofiltered limits.
Best Answer
It should be noted that already in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)} \colon \mathbf{Set} \to \mathbf{Ab}$ does not preserve cofiltered limits. For a cofiltered diagram $D \colon \mathcal I \to \mathbf{Set}$, write $S_i$ for its value at $i \in \mathcal I$, write $S$ for its limit, and write $\pi_i \colon S \to S_i$ for the canonical projection.
There is always a map \begin{align*} \phi \colon \mathbf Z^{(S)} &\to \lim_\leftarrow \mathbf Z^{(S_i)}\\ \sum_{k=1}^n n_k s_k &\mapsto \left( \sum_{k=1}^n n_k\pi_i(s_k) \right)_i. \end{align*} Denote its $i^{\operatorname{th}}$ component by $\phi_i$. For a set $X$ and an element $z=\sum_{k=1}^n n_k x_k \in \mathbf Z^{(X)}$ with $x_k \neq x_{k'}$ for $k \neq k'$, write $\operatorname{Supp}(z)$ for $\{x_1,\ldots,x_n\}$, and denote by $|z|$ its cardinality. If $f \colon X \to Y$ is a map, we denote the induced map $\mathbf Z^{(X)} \to \mathbf Z^{(Y)}$ by $f$ as well (by abuse of notation). For $z \in \mathbf Z^{(X)}$, we have $\operatorname{Supp}(f(z)) \subseteq f(\operatorname{Supp}(z))$, so $|f(z)| \leq |z|$ with equality if and only if $f_* \colon \operatorname{Supp}(z) \to \operatorname{Supp}(f(z))$ is a bijection.
Lemma. The map $\phi$ is injective, and its image consists of those $(x_i)_{i \in \mathcal I}$ for which there exists $n \in \mathbf Z$ with $|x_i| \leq n$ for all $i \in \mathcal I$.
In other words, the image consists of the sequences $(x_i)_i$ of bounded support.
Proof. For injectivity, if $x = \sum_{k=1}^n n_ks_k \in \ker(\phi)$ is such that $s_k \neq s_{k'}$ for $k \neq k'$, then there exists $i \in \mathcal I$ such that $\pi_i(s_k) \neq \pi_i(s_{k'})$ for $k \neq k'$ (here we use that the sum is finite and that $\mathcal I$ is cofiltered). Then $\phi_i(x) = \sum_{k=1}^n n_k\pi_i(s_k)$ is zero by assumption, so all $n_k$ are zero.
For the image, it is clear that $|\phi_i(x)| \leq n$ for all $i \in \mathcal I$ if $x \in \mathbf Z^{(S)}$ has $|x| = n$. Conversely, if $|x_i| \leq n$ for all $i \in \mathcal I$, then decreasing $n$ if necessary, we may assume $|x_{i_0}| = n$ for some $i_0 \in \mathcal I$. Then $|x_i| = n$ for all $f \colon i \to i_0$ since $n = |x_{i_0}| = |D(f)(x_i)| \leq |x_i| \leq n$. Replacing $\mathcal I$ by the coinitial segment $\mathcal I/i_0$ we may therefore assume $|x_i| = n$ for all $i \in \mathcal I$.
Constancy of $|x_i|$ means that for every morphism $f \colon i \to j$ in $\mathcal I$, the map $f_* \colon \operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$ is a bijection. Setting $T = \lim\limits_\leftarrow \operatorname{Supp}(x_i)$, we see that each projection $\pi_i \colon T \to \operatorname{Supp}(x_i)$ is a bijection. Functoriality of the limit gives an injection $T \hookrightarrow S$, giving elements $s_1,\ldots,s_n \in S$ such that $\pi_i(\{s_1,\ldots,s_n\}) = \operatorname{Supp}(x_i)$ for all $i \in \mathcal I$. The coefficients must also be constant under the bijections $\operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$, so we get an element $x = \sum_{k=1}^n n_ks_k \in \mathbf Z^{(S)}$ with $\phi(x) = (x_i)_i$. $\square$
Example. An example where $\phi$ is not surjective: let $S = \mathbf Z_3$ with $S_i = \mathbf Z/3^i$. Define the element $(x_i)_i \in \prod_i \mathbf Z^{(S_i)}$ where $x_i \in \mathbf Z^{(S_i)} = \mathbf Z^{S_i}$ has coordinates (for $k \in \{0,\ldots,3^i-1\}$) given by $$x_{i,k} = \begin{cases} 1, & \text{the first $3$-adic digit of } k \text{ is } 1, \\ -1, & \text{the first $3$-adic digit of } k \text{ is } 2, \\ 0, & k=0.\end{cases}$$ These form an element of $\lim\limits_\leftarrow \mathbf Z^{(S_i)}$: any fibre of $\mathbf Z/3^{i+1} \to \mathbf Z/3^i$ above $k \in \{0,\ldots,3^i-1\}$ consists of $\{k,3^i+k,2 \cdot 3^i+k\}$, of which $x_{i+1,k} = x_{i,k}$ and the others are $1$ and $-1$ since $3^i+k$ starts on $1$ and $2 \cdot 3^i+k$ starts on $2$.
Since $\operatorname{Supp}(x_i) = S_i \setminus \{0\}$, we see that $(x_i)_i$ does not have bounded support, hence is not in the image of $\phi$.
Corollary. For any presheaf topos $\mathbf T = \mathbf{PSh}(\mathscr C) = [\mathscr C^{\operatorname{op}},\mathbf{Set}]$ on a small nonempty category $\mathscr C$, the free functor $\mathbf Z^{(-)} \colon \mathbf T \to \mathbf{Ab}(\mathbf T)$ does not preserve cofiltered limits.
Proof. Let $f \colon \mathscr C \to *$ be the map to the terminal category $*$, on which $\mathbf{PSh}(*) = \mathbf{Set}$. Note that $f$ has a section $g$ since $\mathscr C$ is nonempty. We saw above that in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)}$ does not preserve cofiltered limits. The pullback $f^* \colon \mathbf{Set} \to \mathbf{PSh}(\mathscr C)$ takes a set $S$ to the constant sheaf $\underline{S}$ on $\mathscr C$. Since limits, colimits, and free abelian group objects in presheaf categories are pointwise, $f^*$ commutes with formation of limits, colimits, and free abelian group objects. Thus pulling back everything along $f^*$ gives the natural map $\underline{\mathbf Z}^{(\underline S)} \to \lim\limits_\leftarrow \underline{\mathbf Z}^{(\underline S_i)}$, which is not an isomorphism since it isn't after applying the section $g^*$ (this is just evaluation at an object $g(*) \in \mathscr C$). $\square$
The class of topoi where the free functor does not commute with cofiltered limits is probably much larger still. For instance, my example does not include condensed sets, which is close to a presheaf topos but not quite. I'm not even sure what a topos would look like where these do commute!