# Adjunction between topological spaces and condensed sets

I am trying to prove that the functor \begin{align*} \mathrm{Top} &\longrightarrow \mathrm{Cond}(\mathrm{Set}) \\ X &\longmapsto \underline{X} \end{align*} admits a left adjoint and it is the functor $$T \mapsto T(*)$$ where $$T(*)$$ has the quotient topology of the map $$\bigsqcup_{\underline{S} \rightarrow T} S \rightarrow T(*)$$ where the disjoint union runs over all profinite sets $$S$$ with a map to $$T$$.

To prove this I am trying to construct the following bijection. We have the map \begin{align*} \phi: \mathrm{Hom}(T, \underline{X}) &\longrightarrow \mathrm{Hom}(T(*), X) \\ f &\longmapsto f_* \ . \end{align*} On the other hand, if $$g \in \mathrm{Hom}(T(*), X)$$, then we can consider the map $$\underline{g}: \underline{T(*)} \rightarrow \underline{X}$$. I am trying to construct a map $$i: T \rightarrow \underline{T(*)}$$ to have an inverse for $$\phi$$: \begin{align*} \psi: \mathrm{Hom}(T(*), X) &\longrightarrow \mathrm{Hom}(T, \underline{X}) \\ g &\longmapsto \underline{g} \circ i \ . \end{align*} Is there a way to construct $$i$$? How to prove this adjunction?

Edit: I have managed to construct $$i$$, but I still could not prove that $$\phi$$ and $$\psi$$ are mutually inverses.

I have constructed $$i$$ as the following: For each $$S$$ profinite we have the equivalence $$\mathrm{Hom}(\underline{S},T) = T(S)$$ thanks to the Yoneda Lemma. We can define

\begin{align*} i_S: \mathrm{Hom}(\underline{S},T) &\longrightarrow \mathrm{Hom}(S, T(*)) \\ \eta &\longmapsto \eta_* \ . \end{align*}

It is easy to prove that for $$g \in \mathrm{Hom}(T(*), X)$$ we have $$(\underline{g} \circ i)_* = g$$, but have not managed to prove that for $$f \in \mathrm{Hom}(T, \underline{X})$$ we have $$f = \underline{f_*} \circ i$$. So it only remains to prove that for $$\eta \in \mathrm{Hom}(\underline{S},T)$$ we have

\begin{align*} f_S (\eta) = f_* \circ \eta_* \end{align*}

I suspect this is due to the Yoneda Lemma but I could not prove it.

• Note that for set-theoretic reasons (see Warning 2.14 in Scholze's notes), the functor $\underline T$ is not a condensed set unless $T$ is a T1 topological space. Apr 10, 2022 at 13:39
• An element of $T(S)$ is a map $\underline{S} \to T$. Given such a map we then get a map $S \to \coprod_{\underline{S'}\to T} S' \to T(*)$, which gives an element of $\underline{T(*)}(S)$. That's your counit $T \to \underline{T(*)}$. Apr 10, 2022 at 16:56
• @Wojowu although this problem goes away by working with pyknotic sets instead (after Barwick and Haine), at the expense of assuming Grothendieck universes. Apr 10, 2022 at 17:20
• Thank you, @DylanWilson! I managed to do it! Apr 10, 2022 at 19:36
• @DylanWilson I managed to do it, I have constructed $i$ (the counit). The maps $\phi$ and $\psi$ are inverses? Apr 10, 2022 at 19:51

As Wojowu noted in the comments, one should really look at $$T_1$$ topological spaces. Consider the functors \begin{align*} G\!: \mathbf{Top}_{T_1} &\leftrightarrows \mathbf{Cond}_\kappa:\!F\\ X &\mapsto \big(\underline X \colon S \mapsto \operatorname{Cont}(S,X)\big)\\ T(*) &\leftarrow\!\shortmid T, \end{align*} where $$T(*)$$ is topologised with the quotient topology via the surjection $$\pi \colon \coprod_{(S,f \in T(S))} S \to T(*)\tag{1}\label{1}$$ given on the component $$f \in T(S)$$ by $$f \colon S \to T(*)$$. By the Yoneda lemma, this really means that $$f \colon h_S \to T$$ is a morphism from the representable sheaf $$h_S$$ to $$T$$, and the map $$S \to T(*)$$ is the set-theoretic map $$f_{\{*\}} \colon h_S(*) \to T(*)$$. But there is the consistent abuse of notation to denote $$h_S = \underline S$$ as $$S$$. To see that (\ref{1}) is surjective, use $$S = \{*\}$$.
As Dylan Wilson noted in the comments, the unit $$\eta \colon 1 \to GF$$ is given by the natural transformation \begin{align*} (\eta_T)_S \colon T(S) &\to \operatorname{Cont}(S,T(*)) \\ f &\mapsto \pi \circ \iota_f, \end{align*} where $$\iota_f \colon S \to \displaystyle\coprod_{(S',f')} S'$$ is the insertion of the coordinate corresponding to $$(S,f)$$. This gives the maps \begin{align*} \phi\!: \operatorname{Hom}_{\mathbf{Cond}}(T,\underline X) &\leftrightarrows \operatorname{Cont}(T(*),X) :\!\psi \\ f &\mapsto f_{\{*\}} \\ g \circ \eta_T &\leftarrow\!\shortmid g. \end{align*} It remains to check that these are inverses. If $$g \colon T(*) \to X$$ is continuous, then $$\phi(\psi(g)) \colon T(*) \to X$$ is given by $$\psi(g)_{\{*\}}$$, i.e. the map taking $$t \in T(*)$$ to $$g \circ \pi \circ \iota_f \colon \{*\} \to X$$. But $$\pi \circ g \colon \{*\} \to T(*)$$ is just (the constant map with value) the point $$t$$ (this is how we checked surjectivity of $$\pi$$ earlier!), so $$\psi(g)_{\{*\}}$$ takes $$t$$ to $$g(t)$$, i.e. $$\phi(\psi(g)) = g$$.
Conversely, given a natural transformation $$f \colon T \to \underline X$$, we get another natural transformation $$\psi(\phi(f)) \colon T \to \underline X$$. Write $$g \colon T(*) \to X$$ for $$\phi(f) = f_{\{*\}}$$. If $$S$$ is extremally disconnected and $$h \in T(S)$$, then $$\psi(g)_S$$ takes $$h$$ to the composition $$S \overset{\iota_h}\to \coprod_{(S',h')} S' \overset\pi\to T(*) \overset g\to X.$$ By definition, the composition $$\pi \circ \iota_h \colon S \to T(*)$$ is the map $$h_{\{*\}} \colon h_S(*) \to T(*)$$. Thus $$\psi(g)_S(h)$$ is the composition $$h_S(*) \overset{h_{\{*\}}}\longrightarrow T(*) \overset{f_{\{*\}}}\longrightarrow X.$$ This is the same thing as $$(fh)_{\{*\}} \colon h_S(*) \to \underline X(*)$$, which is the continuous map $$S \to X$$ given by $$f_S(h) \in \underline X(S)$$. We conclude that $$\psi(\phi(f))_S(h) = f_S(h)$$, and since $$S$$ and $$h$$ are arbitrary that $$\psi(\phi(f)) = f$$. $$\square$$
Remark. Morally what's going on here is the following: since $$T(*)$$ has the quotient topology, a map $$g \colon T(*) \to X$$ is continuous if and only if each of the compositions $$gf \colon S \to X$$ are continuous with $$S$$ extremally disconnected and $$f \colon S \to T$$ an $$S$$-point of $$T$$. Thus, a continuous map $$T(*) \to X$$ is the same as continuous maps $$S \to X$$ for every $$S \to T$$, which is roughly what a natural transformation $$T \to \underline X$$ is.
• My understanding is that you don't need it when $\kappa$ is fixed, but a condensed set in Clausen–Scholze is a thing living in the colimit of $\mathbf{Cond}_\kappa$ over 'all' cardinals $\kappa$ (within some universe?), where the transition maps are given by left Kan extension. If I understand correctly, the claim is then that for cardinals $\kappa \leq \lambda$, the left Kan extension of $\underline X \in \mathbf{Cond}_\kappa$ to $\mathbf{Cond}_\lambda$ agrees with $\underline X \in \mathbf{Cond}_\lambda$ when $X$ is $T_1$, but not in general. Aug 12, 2022 at 16:10
• I think it would be useful to make explicit what is the problem with the Kan extension in the particular case when $X=\{s,\eta\}$ is the $2$-point non-$T_1$ space in which only the point $s$ is closed. I don't yet see why the Kan extension disagrees with $\underline{X}$ in this particular case. Aug 12, 2022 at 16:22